PAT 甲级 1054 The Dominant Color (20 分)

1054 The Dominant Color (20 分)

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 [0,2?24??). 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
 1 #include<iostream>
 2
 3 using namespace std;
 4
 5 int main()
 6 {
 7   int M,N,color = -1,num = 1,temp;
 8
 9   cin>>M>>N;
10
11   int motal = M*N;
12
13   for(int i=0;i<motal;++i)
14   {
15     cin>>temp;
16
17     if(color != temp && num > 1)
18     {
19       color = temp;
20       --num;
21     }
22     else if(color != temp && num == 1)
23       color = temp;
24     else
25       ++num;
26   }
27
28   cout<<color<<endl;
29
30   return 0;
31 }

原文地址:https://www.cnblogs.com/cdp1591652208/p/10228021.html

时间: 2025-01-11 23:43:42

PAT 甲级 1054 The Dominant Color (20 分)的相关文章

PAT Advanced 1054 The Dominant Color (20分)

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 mor

PAT:1054. The Dominant Color (20) AC(抓住最多的特点,处理不同和相同的情况,留下剩余的答案)

#include<stdio.h> int main() { int m,n,ans,tmp,times=0; scanf("%d%d",&m,&n); for(int i=0 ; i<n ; ++i) //[思维]题目找出现次数最多的一个,找到不同的,次数减少1,减少到0就换成输入的数字.找到相同的数字,次数+1.最后剩下的一定就是答案 { for(int j=0 ; j<m ; ++j) { scanf("%d",&

PAT:1054. The Dominant Color (20) AC(map法)

#include<stdio.h> #include<map> using namespace std; const int MAX=0x3fffffff; int main() { int m,n; map<int,int> count; //数字与出现次数的map映射 scanf("%d%d",&m,&n); for(int i=0 ; i<n ; ++i) { for(int j=0 ; j<m ; ++j) { i

1054. The Dominant Color (20)

时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 ar

PAT 甲级 1027 Colors in Mars (20 分)(简单,进制转换)

1027 Colors in Mars (20 分) People in Mars represent the colors in their computers in a similar way as the Earth people. That is, a color is represented by a 6-digit number, where the first 2 digits are for Red, the middle 2 digits for Green, and the

PAT (Advanced Level) 1054. The Dominant Color (20)

简单题 #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<queue> #include<algorithm> using namespace std; int a[500000]; int n,m; int main() { scanf("%d%d",&n,&m); for(int i=0;i&

STL_A1054 The Dominant Color (20 分)

https://pintia.cn/problem-sets/994805342720868352/problems/994805422639136768 /* *map的使用 *访问map的键.值 *数字与出现次数的map映射 */ #include<iostream> using namespace std; #include<cstdio> #include<map> int main() { int m,n,color; scanf("%d%d&quo

【PAT甲级】1042 Shuffling Machine (20 分)

题意: 输入洗牌次数K(<=20),输入54张牌每次洗入的位置(不是交换的位置),输出洗好的牌. 代码: #define HAVE_STRUCT_TIMESPEC#include<bits/stdc++.h>using namespace std;multiset<int>st;int a[57];int card[57],b[57];int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

【PAT甲级】1041 Be Unique (20 分)(多重集)

题意: 输入一个正整数N(<=1e5),接下来输入N个正整数.输出第一个独特的数(N个数中没有第二个和他相等的),如果没有这样的数就输出"None". 代码: #define HAVE_STRUCT_TIMESPEC#include<bits/stdc++.h>using namespace std;multiset<int>st;int a[100007];int main(){ ios::sync_with_stdio(false); cin.tie(