首先发三个题目，其链接如下：
/contest/mstestdec3/problems
解析：
第一个题目是暴力，简单的枚举就可以解决，代码如下所示：
#include <stdlib.h>#include <iostream>#include <vector>#include <string>#include <algorithm>using namespace std;char Map[300][300];int Row, Col;char Sen[3][3];vector<pair<int, int> > result;bool Handle();bool Check(int i, int j);void Rotate();int main(){while(Handle()){//empty while}}bool CMP(const pair<int, int>& lhs, const pair<int, int>& rhs);bool Handle(){if(!(cin>>Row>>Col)){return false;}result.clear();for(int i= 0; i< Row; ++i){for(int j= 0; j< Col; ++j){cin>>Map[i][j];}}for(int i= 0; i< 3; ++i){for(int j= 0; j< 3; ++j){cin>>Sen[i][j];}}for(int l= 0; l< 4; ++l){for(int i= 1; i< Row-1; ++i){for(int j= 1; j< Col-1; ++j){if(Check(i,j)){result.push_back(make_pair(i+1,j+1));}}}Rotate();}sort(result.begin(),result.end());result.erase(unique(result.begin(),result.end()),result.end());for(int i= 0; i< result.size(); ++i){cout<<result[i].first<<' '<<result[i].second<<endl;}return true;}bool Check(int i, int j){for(int x= 0; x< 3; ++x){for(int y= 0; y< 3; ++y){if(Sen[x][y]!= Map[i-1+x][j-1+y])return false;}}return true;}void Rotate(){for(int i= 0; i< 3; ++i){for(int j= 0; j<= i; ++j){swap(Sen[i][j],Sen[j][i]);}}for(int i= 0; i< 3; ++i){swap(Sen[0][i],Sen[2][i]);}}bool CMP(const pair<int, int>& lhs, const pair<int, int>& rhs){if(lhs.first!= rhs.first){return lhs.first< rhs.first;}return lhs.second< rhs.second;}
第二个是二分搜索，代码如下所示：
#include <iostream>using namespace std;typedef long long lld;int N, K;lld GS[100001];bool Handle();int main(){while(Handle()){//empty while}}bool Check(int m);//返回true,Ho赢，否则，Hi赢bool Handle(){if(!(cin>>N>>K))return false;for(int i= 0; i< N; ++i){cin>>GS[i];}int l= 1, r= K+1, m;while(l<= r){m= (l+r)>>1;if(!Check(m)){l= m+1;}else{r= m-1;}}while(Check(l))--l;cout<<l+1<<endl;return true;}inline lld MIN(lld x, lld y){return x< y ? x:y;}bool Check(int m){//返回true,Ho赢，否则，Hi赢lld Left= 0;int X= 0;for(int i= 0; i< N; ++i){Left+= m;if(GS[i]<Left){++X;}Left-=min(Left, GS[i]);}if(N&1){return X>=(N+1)>>1;}else{return X>(N>>1);}}
第三个是动态规划，比较简单，代码如下所示：
#include <iostream>#include <stdlib.h>#include <string.h>using namespace std;int dp[101][101][51];bool Visit[101][101][51];int GCD[101][101];int M, N;const int MOD= 1000000007;bool Handle();int gcd(int x, int y);int main(){for(int i= 1; i< 101; ++i){for(int j= 1; j<=i; ++j){GCD[i][j]=GCD[j][i]= gcd(j,i);}}while(Handle()){//empty while}}int Fetch(int i, int j, int k);bool Handle(){if(!(cin>>M>>N))return false;memset(Visit,0,sizeof(Visit));memset(dp,0,sizeof(dp));int Sum= 0;cout<<Fetch(M,M,N)<<endl;return true;}int gcd(int x, int y){return !x ? y:gcd(y%x,x);}int Fetch(int i, int j, int k){if(Visit[i][j][k]){//have gotreturn dp[i][j][k];}Visit[i][j][k]= true;if(j==1){dp[i][j][k]= (k==1)&&(i<=1);return dp[i][j][k];}else{dp[i][j][k]= 0;dp[i][j][k]+= Fetch(i,j-1,k);if(i>= j)dp[i][j][k]+= Fetch(i-j,j-1,k/GCD[j][k]);dp[i][j][k]%=MOD;return dp[i][j][k];}}
