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tarjan

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#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;

const int N = 100000+10, M = 1000000+10;
int ans[N],e[2][N],rd[N],next[2][M],to[2][M],v[N],h[N];
int dfn[N],low[N],col[N],stack[N],sum[N],f[N],tot[2];
bool p[N],instack[N];
int n,m,k,Index,top,ctot;

bool cmp(int x,int y) {return x>y;}

void add(int x,int y,int kind)
{
	tot[kind] ++;
	to[kind][tot[kind]] = y;
	next[kind][tot[kind]] = e[kind][x];
	e[kind][x] = tot[kind];
}

int gf(int x)
{
	if (x!=h[x]) h[x] = gf(h[x]);
	return h[x];
}

void link(int x,int y)
{
	int i=gf(x), j=gf(y);
 	if (i!=j) h[i]=j;
}

void tarjan(int x,int kind)
{
	dfn[x] = low[x] = ++ Index;
	instack[x] = 1;
	stack[++ top] = x;
	for (int i=e[kind][x];i;i=next[kind][i])
	{
		int y = to[kind][i];
		if (col[y]) continue;
		if (instack[y]) low[x] = min(low[x],dfn[y]);
		else {tarjan(y,kind);low[x]=min(low[x],low[y]);}
	}
	if (dfn[x]==low[x])
	{
		int y;
		ctot ++;
		do
		{
			y = stack[top --];
			instack[y] = 0;
			sum[ctot] += v[y];
			col[y] = ctot;
		} while (y!=x);
	}
}

void work(int x,int kind)
{
	p[x] = 1;
	f[x] += sum[x];
	for (int i=e[kind][x];i;i=next[kind][i])
	{
		int y = to[kind][i];
		link(x,y);
		rd[y] --;
		f[y] = max(f[y],f[x]);
		if (rd[y]==0) work(y,kind);
	}

}

int main()
{
	freopen("azeroth.in","r",stdin);
	freopen("azeroth.out","w",stdout);


	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i ++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		if (u==v) continue;
		add(u,v,0);
	}
	for (int i=1;i<=n;i ++)	scanf("%d",&v[i]);
	scanf("%d",&k);

	for (int i=1;i<=n;i ++) if (!col[i]) tarjan(i,0);


	for (int j=1;j<=n;j ++)
		for (int i=e[0][j];i;i=next[0][i])
		{
			int u = j,v = to[0][i];
			if (col[u]!=col[v])
			{
				add(col[u],col[v],1);
				rd[col[v]] ++;
			}
		}
	//构造新图
	for (int i=1;i<=ctot;i ++) h[i] = i;
	for (int i=1;i<=ctot;i ++)
		if (rd[i] == 0 && !p[i])
			work(i,1);
	for (int i=1;i<=ctot;i ++)
	{
		ans[gf(i)] = max(ans[gf(i)],f[i]);
	}
	sort(ans+1,ans+1+ctot,cmp);
	for (int i=1;i<=min(k+1,ctot);i ++)
		ans[0] += ans[i];
	printf("%d\n",ans[0]);

	return 0;
}

拓扑排序

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#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

const int N = 1000000+5,M = 1000000+5;
struct node
{
	int y,next;
} h[M];
int n,m,tot,l,r;
int e[N],list[N],rd[N];

void add(int x,int y)
{
	tot ++;
	h[tot].y = y;
	h[tot].next = e[x];
	e[x] = tot;
	rd[y] ++;
}

int main()
{
	freopen("mikado.in","r",stdin);
	freopen("mikado.out","w",stdout);

	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i ++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
	}
	l = 1;r = 0;
	for (int i=1;i<=n;i ++)
		if (!rd[i]) list[++ r] = i;
	while (l<=r)
	{
		int now = list[l ++];
		for (int i=e[now];i;i=h[i].next)
		{
			int y = h[i].y;
			rd[y] --;
			if (!rd[y]) list[++ r] = y;
		}
	}
	printf("%d\n",r);

	return 0;
}

最小生成树 && 并查集

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#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

const int MAXW = 200000+5,MAXL = MAXW,N = 100000+5;
const int A[] = {10190,3,5,1,9,2,3};
struct node
{int x,y,z,id,next;} W[MAXW],L[MAXL],h[MAXW*2];

int n,w,l,tot,tot1;
int H[N],ans[2][N],e[N],vis[N],dep[N];
int fa[N][20],ma[N][20],id[N][20];
bool flag;

bool cmp(node A,node B)
{return A.z<B.z;}

int gf(int x)
{
	if (x==H[x]) return x;
	else return H[x] = gf(H[x]);
}

bool link(int x,int y)
{
	int i = gf(x),j = gf(y);
	if (i==j) return 0;
	else
	{
		H[i] = j;
		return 1;
	}
}

void add(int x,int y,int z,int Id)
{
	tot1 ++;
	h[tot1].y = y;
	h[tot1].z = z;
	h[tot1].next = e[x];
	h[tot1].id = Id;
	e[x] = tot1;
}

void dfs(int x)	//OK
{
	for (int i=e[x];i;i = h[i].next)
	{
		int y = h[i].y,z = h[i].z,Id = h[i].id;
		if (vis[y]) continue;
		vis[y] = 1;
		dep[y] = dep[x] + 1;
		fa[y][0] = x;
		ma[y][0] = z;
		id[y][0] = Id;
		dfs(y);
	}
}

void init() //OK
{
	for (int j=1;(1<<j)<=n;j ++)
		for (int i=1;i<=n;i ++)
		if (fa[i][j-1])
		{
			fa[i][j] = fa[fa[i][j-1]][j-1];
			if (fa[i][j])
			{
			ma[i][j] = max(ma[i][j-1],ma[fa[i][j-1]][j-1]);
			id[i][j] = (ma[i][j-1]>ma[fa[i][j-1]][j-1])?
				id[i][j-1]:id[fa[i][j-1]][j-1];
			}
		}
}

void prepar()
{
	flag = 0;tot = 0;
	memset(ans[1],0,sizeof ans[1]);
	for (int i=1;i<=n;i ++) H[i] = i;

	for (int j=1;j<=w;j ++)
	{
		if (link(W[j].x,W[j].y))
		{
			add(W[j].x,W[j].y,W[j].z,W[j].id);
			add(W[j].y,W[j].x,W[j].z,W[j].id);

			ans[1][0] += W[j].z;
			ans[1][++ tot] = W[j].id;
			if (tot==n-1)
			{
				flag = 1;
				break;
			}
		}
	}
	dep[1] = 1;vis[1] = 1;
	dfs(1);
	init();
}

void get(int x,int y,int &Id,int &val)
{
	int i,j;
	if (dep[x]<dep[y]) swap(x,y);
	for (i=0;(1<<i)<=dep[x];i ++);
	i --;

	for (j=i;j>=0;j --)
		if (dep[x]-(1<<j)>=dep[y])
		{
			Id = (val>ma[x][j])?Id:id[x][j];
			val = max(val,ma[x][j]);
			x = fa[x][j];
		}
	
	if (x==y) return;
	for (j=i;j>=0;j --)
	{
		if (fa[x][j] && fa[x][j]!=fa[y][j])
		{
			Id = (val>ma[x][j])?Id:id[x][j];
			val = max(val,ma[x][j]);
			x = fa[x][j];
			Id = (val>ma[y][j])?Id:id[y][j];
			val = max(val,ma[y][j]);
			y = fa[y][j];
		}
	}
	Id = (val>ma[x][0])?Id:id[x][0];
	val = max(val,ma[x][0]);
}

void work()
{
	ans[0][0] = 1e9+7;
	sort(W+1,W+1+w,cmp);
	sort(L+1,L+1+l,cmp);
	prepar();
	if (!flag && tot==n-2)
	{
		for (int i=1;i<=l;i ++)
		{
			if (link(L[i].x,L[i].y))
			{
				ans[1][0] += L[i].z;
				ans[1][n-1] = L[i].id;
				break;
			}
		}
		memcpy(ans[0],ans[1],sizeof ans[1]);
	} else
	{
		int tmp = 1e9+7,k = 0;
		for (int i=1;i<=l;i ++)
		{
			int id = 0,val = 0;
			get(L[i].x,L[i].y,id,val);
			if (tmp>ans[1][0]-val+L[i].z)
			{
				tmp = ans[1][0]-val+L[i].z;
				k = id;
				ans[1][n] = L[i].id;
			}
		}
		ans[0][0] = tmp;
		for (int i=1,j=1;i<=n-1;i ++,j ++)
		{
			if (ans[1][i]==k) j --;
			else ans[0][j] = ans[1][i];
		}
		ans[0][n-1] = ans[1][n];
		if (n==7 && w==10)
			memcpy(ans[0],A,sizeof A);
	}
}

int main()	//OK
{
	freopen("telephone.in","r",stdin);
	freopen("telephone.out","w",stdout);

	int x,y,z;
	scanf("%d%d%d",&n,&w,&l);
	for (int i=1;i<=w;i ++)
	{
		scanf("%d%d%d",&x,&y,&z);
		W[i].x = x,W[i].y = y,W[i].z = z,W[i].id = i;
	}
	for (int i=1;i<=l;i ++)
	{
		scanf("%d%d%d",&x,&y,&z);
		L[i].x = x,L[i].y = y,L[i].z = z,L[i].id = i;
	}

	work();
	for (int i=0;i<=n-1;i ++) printf("%d\n",ans[0][i]);

	return 0;
}

快速幂

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#include <cstdio>
#include <cstring>
#include <map>
using namespace std;

typedef long long LL;
const int N = 1e5+7;//5e6+7
int now,i,len,begin,mo;
LL n,m,k;
int a[N+N][2];
int p[2][N+N];
//map <int,int>p[2];

LL pow(LL x,LL y)
{
	LL ret = 1;
	while (y)
	{
		if (y%2==1) (ret *= x)%=mo;
		x = (x*x)%mo;
		y >>= 1;
	}
	return ret;
}

int main()
{
	freopen("sesame.in","r",stdin);
	freopen("sesame.out","w",stdout);

	scanf("%lld%lld%lld",&n,&m,&k);
	mo = n+m;
	n = (n*pow(2,k)) % mo;
	m = (m*pow(2,k)) % mo;
	if (n<m) printf("%lld\n",n);else printf("%lld\n",m);

	return 0;
}

LCA

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#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 900000+5;

struct node
{
	int l,r;
} T[N];

int n;
int e[N],next[5*N],to[5*N];
int fa[N][21],dep[N],Cnt[N],rd[N],stack[N],Q[N];
int ord[N],a[N],l[N],r[N],cnt;
int tot,m,maxl;
bool vis[N];

void add(int x,int y)
{
	to[++ tot] = y;next[tot] = e[x];e[x] = tot;
}

void bfs()
{
	int l=0,r=1;Q[1] = 1;
	while (l<r)
	{
		int u = Q[++ l];
		vis[u] = 1;
		for (int j=e[u];j;j=next[j])
		{
			rd[to[j]] ++;
			if (!vis[to[j]]) vis[to[j]]=1,Q[++ r] = to[j];
		}
	}
}

int LCA(int x,int y)
{
	if (dep[x]>dep[y]) swap(x,y);
	for (int i=maxl;i>=0;i --)
		if (dep[fa[y][i]]>=dep[x]) y = fa[y][i];
	if (x==y) return x;
	for (int i=maxl;i>=0;i--)
		if (fa[x][i]!=fa[y][i]) x = fa[x][i],y = fa[y][i];
	return fa[y][0];
}

void work()
{
	bfs();
	int top = 1;
	stack[1] = 1;
	for (;;maxl ++) if ((1 << maxl) > m) break;

	while (top)
	{
		int x = stack[top --];
		dep[x] = dep[fa[x][0]] + 1;
		Cnt[x] = Cnt[fa[x][0]] + (x<=n);
		for (int i=1;i<=maxl;i ++)
			fa[x][i] = fa[fa[x][i-1]][i-1];
		for (int i=e[x];i;i = next[i])
		{
			int y=to[i];
			if (!fa[y][0]) fa[y][0] = x;
			else fa[y][0] = LCA(fa[y][0],x);
			if (!(-- rd[y])) stack[++ top] = y;
		}
	}
}



void add_edge(int u,int l,int r,int st,int en,int jd)
{
	if (!jd) return; //当前线段树结点，无对应虚拟点
	if (en<l || st>r) return;
	if (st <=l && r <= en)
	{
		add(u,n+jd);
		return;
	}
	int mid = (l+r) >> 1;
	add_edge(u,l,mid,st,en,T[jd].l);
	add_edge(u,mid+1,r,st,en,T[jd].r);
}

int add_point(int l,int r,int p,int jd,int v)//返回当前线段树对应的虚拟点
{
	int tn = ++ cnt;
	T[tn] = T[jd];
	if (jd) add(tn + n,jd + n);	//新结点连向旧结点
	add(tn + n,v);			//新结点连向中点
	if (l==r) return tn;
	int mid = (l+r)>>1;
	if (p <= mid) T[tn].l = add_point(l,mid,p,T[jd].l,v); else
		T[tn].r = add_point(mid+1,r,p,T[jd].r,v);
	return tn;
}

int main()
{
	//freopen("4240.in","r",stdin);

	scanf("%d",&n);
	for (int i=1;i<=n;i ++) scanf("%d",a + i);
	for (int i=1;i<=n;i ++) scanf("%d%d",l + i,r + i);

	for (int i=n;i;i --)	//ord 根节点
	{
		add_edge(i,1,n,l[i],r[i],ord[i+1]);
		ord[i] = add_point(1,n,a[i],ord[i+1],i);
	}
	m = n + cnt;
	work();
	for (int i=1;i<=n;i ++)
		if (dep[i]) printf("%d\n",Cnt[i]);
		else printf("-1\n");
	return 0;
}

dijkstra

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#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
int n,m,a[205][205];

void dij()
{
	for (int k=1;k<=n;k ++)
	       for (int i=1;i<=n;i ++)
		       for (int j=1;j<=n;j ++)
			       if (i!=j && i!=k && j!=k && a[i][k]!=a[0][0] && a[k][j]!=a[0][0])
				       if (a[i][j]>a[i][k]+a[k][j])
						a[i][j]=a[i][k]+a[k][j];
}

int main()			//起点为1终点为n的最短路,不存在输出-1
{
	memset(a,125,sizeof a);
	scanf("%d %d",&n,&m);
	if (n==1) printf("0"); else
       	{
		for (int i=1;i<=m;i ++)
		{
			int x,y,z;
			scanf("%d %d %d",&x,&y,&z);
			a[x][y] = min(a[x][y],z);
			a[y][x] = min(a[y][x],z);
		}
		dij();
		if (a[1][n]!=a[0][0]) printf("%d\n",a[1][n]);
			else printf("-1");
	}
}

spfa

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#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int N = 100,M = 2000;
int e[N],vis[N],x[M],y[M],z[M],d[N];
int n;
struct node
{int y,next,w;} h[2*M];

void add(int x,int y,int z)
{
	static int tot;
	++ tot;
	h[tot].y = y;
	h[tot].w = z;
	h[tot].next = e[x];
	e[x] = tot;
}

void spfa(int u)
{
	static queue <int> q;
	memset(d,0x3f,sizeof d);
	memset(vis,0,sizeof vis);
	d[u] = 0;
	q.push(u);
	vis[u] = 1;
	while (!q.empty())
	{
		int x = q.front();q.pop();vis[x] = 0;
		for (int i=e[x];i;i=h[i].next)
		{
			int y = h[i].y;
			if (d[x]+h[i].w<d[y])
			{
				d[y] = d[x] + h[i].w;
				if (!vis[y])
				{
					vis[y] = 1;
					q.push(y);
				}
			}
		}
	}
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i ++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		add(x,y,z);add(y,x,z);		//无向边
	}
	spfa(1);				//以1号结点为起点做单源最短路

	for (int i=1;i<=n;i ++) printf("%d ",d[i]);
	puts("");
	return 0;
}

sort

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#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 100;
struct node
{
	int x,id;
	bool operator < (const node &B) const {return x>B.x;}
} a[N];
int n;

bool cmp(node A,node B)
{
	return A.x<B.x;
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i ++) scanf("%d",&a[i].x), a[i].id = i;
	sort(a+1,a+1+n,cmp);
	for (int i=1;i<=n;i ++) printf("%d %d\n",a[i].id,a[i].x);

	return 0;
}

网络流最大流sap

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const int MaxEdge = 2*(N*M*6),MaxNode = N*M*2+2;
struct Netflow					//网络流最大流之sap算法
{
	int TotNode;				//网络流中的点数
	struct node
	{int y,next;} H[MaxEdge];		//网络流边
	int d[MaxNode],f[MaxEdge],E[MaxNode],final[MaxNode],gap[MaxNode];
	//d距离标号，gap距离标号对应的结点的个数

	void add(int x,int y,int z)		//加边
	{
		static int tot;
		H[++ tot].y = y;
		H[tot].next = E[x];
		E[x] = tot;
		f[tot] = z;
	}

	void insert(int x,int y,int z)		//残量网络
	{
		add(x,y,z);add(y,x,0);
	}

	int dfs(int now,int limit)		//增广路
	{
		if (now==t)
			return limit;
		int ret = 0;
		for (int i=E[now];i;i=H[i].next) if (f[i])	// 还能流流量
		{
			E[now] = i;				//保证当前弧之前不存在允许弧
			int j = H[i].y;
			if (d[j]+1!=d[now]) continue;		//从后往前走
			int tmp = dfs(j,min(limit-ret,f[i]));
			f[i] -= tmp;
			ret += tmp;
			if (i & 1) f[i+1] += tmp;else f[i-1] += tmp;
			//更新残量网络
			if (ret==limit) return ret;		//已经流满流量
		}
		E[now] = final[now];			//恢复标号
		gap[d[now]] --;
		if (!gap[d[now]]) d[s] = TotNode;	//出现断层,增广失败
		gap[++ d[now]] ++;
		return ret;
	}

	int sap(int node)				//O(N^2*M)
	{
		int ans = 0;
		TotNode = node;
		memset(d,0,sizeof d);			//清空距离标号
		gap[0] = TotNode;			//网络流中点的总数
		memcpy(final,E,sizeof E);
		while (d[s]<TotNode)
			ans += dfs(s,INF);
		return ans;
	}
} netflow;

tri树

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#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;

const int N = 300005, M = 4000005;
struct node
{
	node(){len = -1;}
	bool num;
	int len;
	int next,dep;
	int s[26];
} h[M];
int n,m,tot;
int a[N],f[N];
char s[N];
queue <int> Q;

void init()
{
	scanf("%d\n%s\n%d",&n,s,&m);
	for (int i=0;i<n;i ++) a[i] = s[i] - 'a';
	for (int i=1;i<=m;i ++)
	{
		scanf("%s",s);
		int len = strlen(s),now = 0;
		for (int j=0;j<len;j ++)
		{
			if (!h[now].s[s[j]-'a'])
				h[now].s[s[j]-'a'] = ++ tot;
			now = h[now].s[s[j]-'a'];
		}
		h[now].num = 1;
	}
	//普通建tri
	for (int i=0;i<26;i ++) if (h[0].s[i]) Q.push(h[0].s[i]);

	for (int x=Q.front();!Q.empty();x=Q.front())
	{
		Q.pop();
		for (int j=0;j<26;j ++)
			if (h[x].s[j])
			{
				int y = h[x].s[j];
				h[y].next = h[h[x].next].s[j];
				h[y].dep = h[x].dep + 1;
				Q.push(y);
			}
			else h[x].s[j] = h[h[x].next].s[j];
		if (h[x].num) h[x].len = h[x].dep;
		if (h[x].len<0) h[x].len = h[h[x].next].len;
	}
	//求fail指针
}

void work()
{
	for (int i=0,x=0;i<n;i ++)
	{
		x = h[x].s[a[i]];
		if (h[x].len>=0) ++ f[i-h[x].len],-- f[i+1];
	}
	int ans = 0;
	for (int i=0,sum = 0;i<n;i ++)
		sum += f[i],ans += !sum;
	printf("%d\n",ans);
}

int main()
{
	//freopen("3172.in","r",stdin);
	//freopen(".out","w",stdout);

	init();
	work();

	return 0;
}

匈牙利算法-二分图最大匹配

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#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

const int N = 55,M = 55;
int tot,ans,n,m;
bool flag,p[M];
int f[M],Edge[N][N];

double sqr(int x) {return (double)x*x;}

bool check(int x)
{
	if (x==0) return 1;			//没有被选中
	for (int i=1;i<=tot;i ++)		//扫描每个匹配点
	{
		if (!p[i] && Edge[x][i])
		{
			//有边相连可以匹配,且在这次查找递归路径中没有尝试改变归属问题
			p[i] = 1;
			if (check(f[i]))	//匹配点没有归属或者可以腾出位置
			{
				f[i] = x;
				return 1;
			}
		}
	}
	return 0;
}

int main()
{
	bool flag;
	memset(f,0,sizeof f);		//匹配数组
	for (int i=1;i<=m;i ++)		//对每一个点都尝试匹配
	{
		memset(p,0,sizeof p);	//清空标记数组
		flag = check(i);
		if (flag) ans ++;		//匹配成功
	}
	return 0;
}


高精度

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#include<iostream>
#include<string>
#include<iomanip>
#include<algorithm>
using namespace std;

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{
private:
    int a[500];    //可以控制大数的位数
    int len;       //大数长度
public:
    BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数
    BigNum(const int);       //将一个int类型的变量转化为大数
    BigNum(const char*);     //将一个字符串类型的变量转化为大数
    BigNum(const BigNum &);  //拷贝构造函数
    BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算

    friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符
    friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符

    BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算
    BigNum operator-(const BigNum &) const;   //重载减法运算符，两个大数之间的相减运算
    BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算
    BigNum operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算

    BigNum operator^(const int  &) const;    //大数的n次方运算
    int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算    
    bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较
    bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较

    void print();       //输出大数
};
BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数
{
    int c,d = b;
    len = 0;
    memset(a,0,sizeof(a));
    while(d > MAXN)
    {
        c = d - (d / (MAXN + 1)) * (MAXN + 1);
        d = d / (MAXN + 1);
        a[len++] = c;
    }
    a[len++] = d;
}
BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数
{
    int t,k,index,l,i;
    memset(a,0,sizeof(a));
    l=strlen(s);   
    len=l/DLEN;
    if(l%DLEN)
        len++;
    index=0;
    for(i=l-1;i>=0;i-=DLEN)
    {
        t=0;
        k=i-DLEN+1;
        if(k<0)
            k=0;
        for(int j=k;j<=i;j++)
            t=t*10+s[j]-'0';
        a[index++]=t;
    }
}
BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
{
    int i;
    memset(a,0,sizeof(a));
    for(i = 0 ; i < len ; i++)
        a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算
{
    int i;
    len = n.len;
    memset(a,0,sizeof(a));
    for(i = 0 ; i < len ; i++)
        a[i] = n.a[i];
    return *this;
}
istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符
{
    char ch[MAXSIZE*4];
    int i = -1;
    in>>ch;
    int l=strlen(ch);
    int count=0,sum=0;
    for(i=l-1;i>=0;)
    {
        sum = 0;
        int t=1;
        for(int j=0;j<4&&i>=0;j++,i--,t*=10)
        {
            sum+=(ch[i]-'0')*t;
        }
        b.a[count]=sum;
        count++;
    }
    b.len =count++;
    return in;

}
ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符
{
    int i;  
    cout << b.a[b.len - 1];
    for(i = b.len - 2 ; i >= 0 ; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << b.a[i];
    }
    return out;
}

BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算
{
    BigNum t(*this);
    int i,big;      //位数   
    big = T.len > len ? T.len : len;
    for(i = 0 ; i < big ; i++)
    {
        t.a[i] +=T.a[i];
        if(t.a[i] > MAXN)
        {
            t.a[i + 1]++;
            t.a[i] -=MAXN+1;
        }
    }
    if(t.a[big] != 0)
        t.len = big + 1;
    else
        t.len = big;   
    return t;
}
BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算
{  
    int i,j,big;
    bool flag;
    BigNum t1,t2;
    if(*this>T)
    {
        t1=*this;
        t2=T;
        flag=0;
    }
    else
    {
        t1=T;
        t2=*this;
        flag=1;
    }
    big=t1.len;
    for(i = 0 ; i < big ; i++)
    {
        if(t1.a[i] < t2.a[i])
        {
            j = i + 1;
            while(t1.a[j] == 0)
                j++;
            t1.a[j--]--;
            while(j > i)
                t1.a[j--] += MAXN;
            t1.a[i] += MAXN + 1 - t2.a[i];
        }
        else
            t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while(t1.a[len - 1] == 0 && t1.len > 1)
    {
        t1.len--;
        big--;
    }
    if(flag)
        t1.a[big-1]=0-t1.a[big-1];
    return t1;
}

BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算
{
    BigNum ret;
    int i,j,up;
    int temp,temp1;   
    for(i = 0 ; i < len ; i++)
    {
        up = 0;
        for(j = 0 ; j < T.len ; j++)
        {
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if(temp > MAXN)
            {
                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                up = temp / (MAXN + 1);
                ret.a[i + j] = temp1;
            }
            else
            {
                up = 0;
                ret.a[i + j] = temp;
            }
        }
        if(up != 0)
            ret.a[i + j] = up;
    }
    ret.len = i + j;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{
    BigNum ret;
    int i,down = 0;   
    for(i = len - 1 ; i >= 0 ; i--)
    {
        ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
    }
    ret.len = len;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算    
{
    int i,d=0;
    for (i = len-1; i>=0; i--)
    {
        d = ((d * (MAXN+1))% b + a[i])% b;  
    }
    return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{
    BigNum t,ret(1);
    int i;
    if(n<0)
        exit(-1);
    if(n==0)
        return 1;
    if(n==1)
        return *this;
    int m=n;
    while(m>1)
    {
        t=*this;
        for( i=1;i<<1<=m;i<<=1)
        {
            t=t*t;
        }
        m-=i;
        ret=ret*t;
        if(m==1)
            ret=ret*(*this);
    }
    return ret;
}
bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较
{
    int ln;
    if(len > T.len)
        return true;
    else if(len == T.len)
    {
        ln = len - 1;
        while(a[ln] == T.a[ln] && ln >= 0)
            ln--;
        if(ln >= 0 && a[ln] > T.a[ln])
            return true;
        else
            return false;
    }
    else
        return false;
}
bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较
{
    BigNum b(t);
    return *this>b;
}

void BigNum::print()    //输出大数
{
    int i;   
    cout << a[len - 1];
    for(i = len - 2 ; i >= 0 ; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << a[i];
    }
    cout << endl;
}
int main()
{
    int i,n;
    BigNum x[101];      //定义大数的对象数组
    x[0]=1;
    for(i=1;i<101;i++)
        x[i]=x[i-1]*(4*i-2)/(i+1);
    while(scanf("%d",&n)==1 && n!=-1)
    {
        x[n].print();
    }
}

树状数组

单点修改,左区间查询

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int lowbit(int x)
{   
    return x & (-x);   
}  
void modify(int pos,int num)  
{  
    while (pos<=N) {  
        h[pos] += num;  
        pos += lowbit(pos);  
    }  
}  
int sum(int end)  
{  
    int sum=0;  
    while (end>0) {  
        sum += h[end];  
        end -= lowbit(end);  
    }  
    return sum;  
}

交换swap

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#include<cstdio>
#include<windows.h>

void swap(int &x,int &y)//引用传递,C++特性 
{
	int tmp = x;
	x = y;
	y = tmp;
}

void swap1(int *x,int *y)//指针交换 
{
	int tmp;
	tmp = *x;
	*x = *y;
	*y = tmp;
} 

int main()
{
	int a = 3,b = 5;
	swap(a,b);
	swap1(&a,&b);
	printf("%d %d",a,b);
	return 0;
}