HDOJ-考試排名

沒什麼難度，只要能處理一些字串就行了

格式很蛋疼啊PE了兩次

 

#include <algorithm><br />#include <cstdio><br />#include <cstring><br />#include <iostream><br />using namespace std;<br />struct Node{<br />char name[20];<br />int cnt, time;<br />}a[50000];<br />bool cmp(Node p, Node q){<br />if(p.cnt != q.cnt)<br />return p.cnt > q.cnt;<br />if(p.time != q.time)<br />return p.time < q.time;<br />return strcmp(p.name, q.name) < 0;<br />}<br />int main(){<br />int nproblem, k, i = 0, j, penalty, x, y;<br />char s[100];<br />cin >>nproblem >>penalty;<br />while(cin >>a[i].name){<br />if(a[i].name[0] == '#')break;<br />a[i].cnt = a[i].time = 0;<br />for(j = 0; j < nproblem; j++){<br />cin >>s;<br />if(s[0] == '0' || s[0] == '-')<br />continue;<br />a[i].cnt++;<br />for(k = 0; s[k]; k++){<br />if(s[k] == '(')<br />break;<br />}<br />if(s[k]){<br />x = y = 0;<br />for(k = 0; s[k] != '('; k++){<br />x = x * 10 + s[k] - '0';<br />}<br />for(k++; s[k] != ')'; k++){<br />y = y * 10 + s[k] - '0';<br />}<br />a[i].time += x + y * penalty;<br />}else{<br />x = 0;<br />for(k = 0; s[k]; k++){<br />x = x * 10 + s[k] - '0';<br />}<br />a[i].time += x;<br />}<br />}<br />i++;<br />}<br />sort(a,a+i,cmp);<br />for(j = 0; j < i; j++){<br />printf("%-10s %2d %4d/n",a[j].name,a[j].cnt, a[j].time);<br />}<br />return 0;<br />}