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来源：互联网发布：揭阳网络广告公司编辑：程序博客网时间：2024/06/16 04:27

思路：统计字符的最大值若有相同的以字典序第一个为准输入
简单题

#include<iostream>#include<cstdio>#include<string>#include<algorithm>using namespace std;int main(){    int T, m, n;    int i, j;    int a_count = 0, g_count = 0,c_count=0,t_count=0;    char DNA[50][10001];    char output[10001];    int min;    cin >> T;    while (T--)    {        min = 0;        cin >> m >> n;        getchar();        for (i = 0; i < m; i++)                     //输入序列            fgets(DNA[i], n+2, stdin);        for (i = 0; i < n; i++)                     //统计个数        {            for (j = 0; j < m; j++)            {                switch (DNA[j][i])                {                case 'A':                    a_count++;                    break;                case 'G':                    g_count++;                    break;                case 'C':                    c_count++;                    break;                case 'T':                    t_count++;                    break;                }            }            if (a_count >= c_count&&a_count >= g_count &&a_count >= t_count)            {                output[i] = 'A';                min += (m - a_count);            }            if (c_count >= g_count &&c_count >= t_count&&c_count > a_count)            {                output[i] = 'C';                min += (m - c_count);            }            if (g_count >= t_count&&g_count > a_count&&g_count > c_count)            {                output[i] = 'G';                min += (m - g_count);            }            if (t_count > a_count&&t_count > g_count&&t_count > c_count)            {                output[i] = 'T';                min += (m - t_count);            }            a_count = 0;            g_count = 0;            c_count = 0;            t_count = 0;        }        for (i = 0; i < n; i++)            printf("%c", output[i]);        printf("\n%d\n", min);    }    return 0;}

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