﻿PAT(A) 1054. The Dominant Color (20)-Ocrosoft

# PAT(A) 1054. The Dominant Color (20)

Behind the scenes in the computer’s memory, color is always talked about as a series of 24 bits of information for each pixel. In an image, the color with the largest proportional area is called the dominant color. A strictly dominant color takes more than half of the total area. Now given an image of resolution M by N (for example, 800×600), you are supposed to point out the strictly dominant color.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: M (<=800) and N (<=600) which are the resolutions of the image. Then N lines follow, each contains M digital colors in the range [0, 224). It is guaranteed that the strictly dominant color exists for each input image. All the numbers in a line are separated by a space.

Output Specification:

For each test case, simply print the dominant color in a line.

Sample Input:

```5 3
0 0 255 16777215 24
24 24 0 0 24
24 0 24 24 24
```

Sample Output:

```24
```
string会超时，用int+scanf；

```#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
#define strend string::npos
#define ms(a) memset(a,0,sizeof(a))
#define rep(a,v,b) for(int a=v;a<b;a++)
#define repe(a,v,b) for(int a=v;a<=b;a++)
#define pre(a,v,b) for(int a=v;a>b;a--)
#define pree(a,v,b) for(int a=v;a>=b;a--)
#define lowbit(x) x&-x
typedef long long LL;
const LL LINF = LLONG_MAX / 2;
const int INF = INT_MAX / 2;
const int MAXN = 1e4 + 10;
const int MOD = 1000000007;
int gcd(int a, int b)
{
if (!b)return a;
return gcd(b, a%b);
}
/*(◕‿‿◕)(◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕)*/
/*(◕‿‿◕) 签订契约,成为马猴烧酒吧 (◕‿‿◕)*/
/*(◕‿‿◕)(◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕)*/
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
map<int, int> mp;
bool flag = 0;
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
int t; scanf("%d", &t);
if (flag)continue;
if (mp.count(t))mp[t]++;
else mp[t] = 1;
if (mp[t] > n*m / 2) { flag = 1; printf("%d\n", t); }
}
}
return 0;
}
```