题目链接
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3961
题意
给出两个人的发消息的记录,然后 如果有两人在连续M天及以上 互发消息,超过几天 加几点 友谊值 最后求 友谊值
思路
先把第一个 连续区间 >= M 的 存下来 然后第二次 输入的时候 去判断。
AC代码
#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
using namespace std;
typedef long long LL;
const double PI = 3.14159265358979323846264338327;
const double E = 2.718281828459;
const double eps = 1e-6;
const int MAXN = 0x3f3f3f3f;
const int MINN = 0xc0c0c0c0;
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
int n, q;
int i, j;
scanf("%d%d", &n, &q);
map <string, string> M;
M.clear();
string s, temp = "";
for (i = 0; i < q; i++)
temp += "0";
int m;
scanf("%d", &m);
for (i = 0; i < m; i++)
{
cin >> s;
M[s] = temp;
}
for (i = 0; i < q; i++)
{
scanf("%d", &m);
for (j = 0; j < m; j++)
{
cin >> s;
M[s][i] = '1';
}
}
// map <string, string>::iterator it;
// for (it = M.begin(); it != M.end(); it++)
// cout << it -> first << " " << it -> second << endl;
string ans;
int num;
for (i = 0; i < n; i++)
{
temp.clear();
for (j = 0; j < q; j++)
{
scanf(" %d", &num);
temp += num + '0';
}
int flag = 0;
ans.clear();
map <string, string>:: iterator it;
for (it = M.begin(); it != M.end(); it++)
{
if (it -> second == temp )
{
if (flag == 1)
{
flag = 0;
break;
}
else
{
flag = 1;
ans = it -> first;
}
}
}
if (flag)
cout << ans << endl;
else
printf("Let's go to the library!!\n");
}
}
}