立志用最少的代碼做最高效的表達

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PAT甲級最優題解——>傳送門

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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

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散列表，永遠滴神

流同步問題：
如果采用cin、cout輸入輸出，需要取消流同步，即：ios::sync_with_stdio(false);。 速度相較于scanf、printf來說更快。

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集合和映射問題：

儲備知識擴展(很重要，提高效率，降低碼量)：

• set——有序去重集合。

• map——有序去重映射

• multiset——有序不去重集合

• multimap——有序不去重映射

• unordered_set——無序不去重集合(普通集合)

• unordered_map——無序不去重映射(普通映射)

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#include<bits/stdc++.h> using namespace std; int main() {ios::sync_with_stdio(false);int n, m; cin >> n >> m;unordered_map<int, int>um;for(int i = 0; i < n*m; i++) {int x; cin >> x; um[x]++;if(um[x] >= (n*m)/2+1) { cout << x; return 0; }}return 0; }

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耗時：

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求贊哦~ (?ω?)

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