PAT 1012（甲级） The Best Rank

坑点：不要把6为的学号当int型储存，要按照字符串储存，否则如果是001101，输入1101，则按照int型可以匹配，测试点第一个就是测试这个。

#include<iostream>#include<string.h>using namespace std;int main(){int N, M;cin >> N >> M;int table[2000][4];//table[i][0],到table[i][3]依次存入总分,C,M,E的成绩char id[2000][7];//学号int rank[2000][4] = { 0 };//每位同学各科排名和总排名int best[2000][2] = { 0 };//最佳排名,和对应的科目for (int i = 0; i < N; i++){int sum = 0;cin >> id[i];id[i][6] = '\0';for (int k = 0; k < 4; k++){if (k == 0)continue;cin >> table[i][k];sum += table[i][k];}table[i][0] = sum;//计算总分存入table[i][0]}for (int i = 0; i < N; i++){int bestrank = -1;int bestcourse;for (int j = 0; j < 4; j++)//j对应科目{int max = 0;for (int k = 0; k < N; k++){if (table[i][j] < table[k][j]){if (max < table[k][j])max = table[k][j];rank[i][j]++;}}if (bestrank > rank[i][j] || bestrank == -1)//每位学生最优秀科目{bestrank = rank[i][j];bestcourse = j;}}best[i][0] = bestrank+1;//rank数组初始化为0，因此每个人的名次要加1best[i][1] = bestcourse;}for (int i = 0; i < M; i++){int k;char nid[7];cin >> nid;nid[6] = '\0';for (k = 0; k < N+1; k++){if(strcmp(nid,id[k])==0)break;}if (k != N + 1){cout << best[k][0] << ' ';switch (best[k][1]){case 0:cout << 'A'; break;case 1:cout << 'C'; break;case 2:cout << 'M'; break;case 3:cout << 'E'; break;}}elsecout << "N/A";cout << endl;}system("pause");return 0;}