Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19296 Accepted Submission(s): 5695
200.00 3 2 A:23.50 B:100.00 1 C:650.00 3 A:59.99 A:120.00 X:10.00 1200.00 2 2 B:600.00 A:400.00 1 C:200.50 1200.50 3 2 B:600.00 A:400.00 1 C:200.50 1 A:100.00 100.00 0
123.50 1000.00 1200.50
这道题神坑。。。有以下几个点需要注意:
1.是一张条上同一大类不能超过600
2.纸条上只要存在不是ABC三类中的一类,都不可以报销,所以第一组数据是123.5
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct node{ double sum; int mark; }note[40]; int cmp(node a,node b){ return a.sum>b.sum; } int main(){ double q; int n; while(~scanf("%lf",&q)){ scanf("%d",&n); if(n==0) break; else{ int m,i,j; double temp,a[3]; char c; for(i=0;i<n;i++){ scanf("%d",&m); note[i].sum=0; note[i].mark=1; memset(a,0,sizeof(a)); for(j=0;j<m;j++){ scanf(" %c:%lf",&c,&temp); if((c!='A')&&(c!='B')&&(c!='C')){ note[i].mark=0; } note[i].sum+=temp; if(c=='A'){ a[0]+=temp; } if(c=='B'){ a[1]+=temp; } if(c=='C'){ a[2]+=temp; } if((a[2]>600)||(a[1]>600)||(a[0]>600)){ note[i].mark=0; } if(note[i].sum>1000){ note[i].mark=0; } } } sort(note,note+n,cmp); double fin=0; for(i=0;i<n;i++){ if(note[i].sum<=q&¬e[i].mark!=0){ fin+=note[i].sum; q-=note[i].sum; } } printf("%.2lf\n",fin); } } return 0; }