A.计蒜客 - T1381

输出hello world
万恶之源

B.51Nod - 2060

全排列输出
不要用STL的next_permutation,会超时

#include <bits/stdc++.h>
using namespace std;
const int maxn=14;
int dt[maxn];
int vis[maxn];int n;
void dfs(int depth)
{if(depth==n){for(int i=0;; ++i){if(i==n)break;if(i==0)cout<<dt[i];else cout<<" "<<dt[i];}cout<<endl;return ;}for(int i=1; i<=n; ++i){if(vis[i]==0){vis[i]=1;dt[depth]=i;dfs(depth+1);vis[i]=0;}}
}
int main()
{cin>>n;dfs(0);return 0;
}

C HDU - 1715

求斐波那契数列（不过比较大到1000）
用数组来存每一位，模拟进制运算

#include<iostream>
#include<cstdio>
#include<cstring>typedef long long LL;
using namespace std;
int f[1005][605];void init()
{f[1][0] = f[2][0] = 1;f[1][1] = f[2][1] = 1;int len = 1;for(int i = 3;i <= 1000;i ++){for(int j = 1;j <= len;j ++){f[i][j] += f[i-1][j] + f[i-2][j];if(f[i][j] >= 10){f[i][j+1] += f[i][j] / 10;f[i][j] %= 10;}}while(f[i][len+1]) len ++;f[i][0] = len;}
}int main()
{int n, t;init();scanf("%d",&t);while(t --){scanf("%d",&n);for(int i = f[n][0];i >= 1;i --)printf("%d", f[n][i]);printf("\n");}return 0;
}

#include <iostream>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
using namespace std;
typedef long long ll;
int main(){int n,q,i,j,temp=0;cin>>q;while(q--){cin>>n;char a[10010]="1",b[10010]="1",c[10010]={0};for(i=0;i<n-2;i++){int len=max(strlen(a),strlen(b));for(j=0;j<len;j++){   //模拟加法temp=0;if(a[j]>='0'){    //短的数不加temp+=a[j]-'0';}if(b[j]>='0'){temp+=b[j]-'0';}if(temp>=10){      //判断进位if(c[j]>='0'){c[j]+=temp-10;}else{c[j]+=temp-10+'0';}c[j+1]=1+'0';}else{if(c[j]>='0'){if(temp==9){   //若前位进位了,而且加上的数字是9,那么还要进位!!!c[j]='0';c[j+1]='1';}else{c[j]+=temp;}}else{c[j]+=temp+'0';}}}strcpy(a, b);strcpy(b, c);memset(c, 0, sizeof(c));}int len=strlen(b);for(i=len-1;i>=0;i--){  //倒着输出printf("%c",b[i]);}printf("\n");}return 0;
}

D POJ - 2251

三维迷宫BFS搜就完事了

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
using namespace std;
const int maxn=32;
struct node
{int x;int y;int z;int step;
} start,fin; 
char a[maxn][maxn][maxn];
bool vis[maxn][maxn][maxn];int dx[]={0,0,1,-1,0,0};
int dy[]={1,-1,0,0,0,0};
int dz[]={0,0,0,0,1,-1};int l,r,c;bool flag=false;
queue<node> q;void bfs()
{while(q.size()) q.pop();memset(vis,false,sizeof(vis));q.push(start);vis[start.x][start.y][start.z]=true;while(!q.empty()){node temp=q.front();q.pop();	if(temp.x==fin.x && temp.y==fin.y && temp.z==fin.z){flag=1;printf("Escaped in %d minute(s).\n",temp.step);return ;}for(int i=0;i<6;i++){node next;next.x=temp.x+dx[i];next.y=temp.y+dy[i];next.z=temp.z+dz[i];if( next.x>=0&&next.x<l && next.y>=0&&next.y<r && next.z>=0&&next.z<c &&!vis[next.x][next.y][next.z] && a[next.x][next.y][next.z]!='#'){vis[next.x][next.y][next.z]=true;next.step=temp.step+1;q.push(next); }		} }
}
int main()
{while(cin>>l>>r>>c){if(l==0&&r==0&&c==0){break;}for(int i=0;i<l;i++){for(int j=0;j<r;j++){for(int k=0;k<c;k++){cin>>a[i][j][k];if(a[i][j][k]=='S'){start.x=i;start.y=j;start.z=k;start.step=0;}if(a[i][j][k]=='E'){fin.x=i;fin.y=j;fin.z=k;fin.step=0;}}}}	flag=0;bfs();	if(!flag){cout<<"Trapped!"<<endl;	} }return 0;
}

E 计蒜客 - T2028

跳石头，经典二分题目， noip原题

#include<bits/stdc++.h>
using namespace std;
long long int a[50004];long long int d,n,m;
bool judge(int x){int tot=0;int i=0;int now=0;while(i<n+1){i++;if(a[i]-a[now]<x)tot++;else now=i;}if(tot>m)return 0;else return 1;
}
int main()
{cin>>d>>n>>m;for(int i=1;i<=n;i++){cin>>a[i];} a[n+1]=d;int l=1;int r=d;int ans;while(l<=r){int mid=(l+r)>>1;if(judge(mid)){l=mid+1;ans=mid;}else r=mid-1;}cout<<ans<<endl;
}

F HDU - 1702

#include<iostream>
#include<cstdio>
#include<queue>
#include<stack>
using namespace std;
void cnd(int sum,bool f)
{if(f==1){//先进先出 queue<int>q;string a;int b;for(int i=1;i<=sum;i++){cin>>a;if(a[0]=='I'){cin>>b;q.push(b);}else {if(q.empty()){cout<<"None"<<endl;}else {cout<<q.front()<<endl;q.pop();}}}while(!q.empty())q.pop();}else{stack<int>s;string a;int b;for(int i=1;i<=sum;i++){cin>>a;if(a[0]=='I'){cin>>b;s.push(b);}else {if(s.empty()){cout<<"None"<<endl;}else {cout<<s.top()<<endl;s.pop();}}}while(!s.empty())s.pop();}}
int main()
{int n;cin>>n;int a;string s;for(int i=1;i<=n;i++){cin>>a>>s;
//		cout<<s[0]<<endl;bool f;if(s[2]=='F')f=1;//先进先出 else f=0;cnd(a,f); }return 0;
}

G POJ - 1011

详细题解

H POJ - 3494

求最大全1子矩阵的面积
题解

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;const int MAXN = 2007;
int n,m;
int h[MAXN],s[MAXN],L[MAXN],R[MAXN];int solve()
{int t = 0;for(int i = 0; i < m; ++i){while(t > 0 && h[s[t-1]] >= h[i])t--;if(t == 0) L[i] = 0;else L[i] = s[t-1]+1;s[t++] = i;}//找最左边 t = 0;for(int i = m-1; i >= 0; --i)//最右边 {while(t > 0 && h[s[t-1]] >= h[i])t--;if(t == 0) R[i] = m-1;else R[i] = s[t-1]-1;s[t++] = i;}int ret = 0;for(int i = 0; i < m;  ++i)ret = max(ret,h[i]*(R[i]-L[i]+1));return ret;
}int main()
{int num;while(~scanf("%d %d",&n,&m)){memset(h,0,sizeof(h));int res = 0;for(int i = 0; i < n; ++i){for(int j = 0; j < m; ++j){scanf("%d",&num);if(num==1)h[j]++;else h[j]=0;}res = max(res,solve());}printf("%d\n",res);}return 0;
}
/*
4 5
0 1 1 0 1
0 1 1 1 1
0 0 1 1 0
0 0 1 1 1
*/ 

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring> int n,m,ans;int map[2333][2333];int h[2333][2333],sta[2333],L[2333],R[2333];//h[i][j]:第i行第j列元素往上最长的连续1长度//维护单调非递减栈void init(int row){int top = 0,tmp;h[row][m+1] = 0;for(int j = 1;j <= m + 1;j ++) {while(1) {tmp = sta[top];if(h[row][tmp] <= h[row][j]) break;//以j作为延伸点，确定边界，保证单调性。 R[tmp] = j;-- top;}L[j] = tmp;sta[++top] = j;}for(int j = 1;j <= m;j ++) {if(map[row][j] == 0) continue;int len = R[j] - L[j] - 1;ans = max(ans,h[row][j] * len);}}int main(){while(scanf("%d%d",&n,&m) != EOF) {for(int i = 1;i <= n;i ++) {for(int j = 1;j <= m;j ++) scanf("%d",&map[i][j]);}memset(h,0,sizeof(h));for(int j = 1;j <= m;j ++) {for(int i = 1;i <= n;i ++) {if(map[i][j] == 1) {h[i][j] = 1;while(map[++i][j] == 1) h[i][j] = h[i-1][j] + 1;-- i;}}}ans = 0;for(int i = 1;i <= n;i ++) init(i);printf("%d\n",ans);}return 0;}