HDU1084

细节题
1，注意输出
2，当a[i].sum有偶数个时，前一半比后一半多5
3，当只有一个时，除100，50以外，输出90，80，70，60等等之类的
View Code
  1 #include<stdio.h>
  2 #include<stdlib.h>
  3 #include<string.h>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<queue>
  7 #include<map>
  8 #include<math.h>
  9 using namespace std;
 10 const int maxn = 105;
 11 const int inf = 0x7fffffff;
 12 struct node{
 13     int sum,time,id,ans;
 14 }a[ maxn ];
 15 bool cmp(node a,node b){
 16     if( a.sum!=b.sum )return a.sum>b.sum;
 17     else return a.time<b.time;
 18 }
 19 bool cmp2(node a,node b){
 20     return a.id<b.id;
 21 }
 22 int main(){
 23     int n;
 24     int n1,n2,n3,n4,n5;
 25     while( scanf("%d",&n)!=EOF ){
 26         if( n<=0 ) break;
 27         char s[ 18 ];
 28         n1=n2=n3=n4=n5=0;
 29         for( int i=1;i<=n;i++ ){
 30             scanf("%d %s",&a[ i ].sum,s);
 31             a[ i ].id=i;
 32             if( a[i].sum==1 ) n1++;
 33             if( a[i].sum==2 ) n2++;
 34             if( a[i].sum==3 ) n3++;
 35             if( a[i].sum==4 ) n4++;
 36             if( a[i].sum==5 ) n5++;
 37             int res=0;
 38             res=( (s[0]-'0')*10+(s[1]-'0') )*3600+( (s[3]-'0')*10+(s[4]-'0') )*60+( (s[6]-'0')*10+(s[7]-'0'));
 39             a[i].time=res;
 40         }
 41         //for( int i=1;i<=n;i++ ){
 42         //    printf("id:%d time:%d sum:%d\n",a[i].id,a[i].time,a[i].sum);
 43         //}
 44     //    printf("\n");
 45         sort( a+1,a+n+1,cmp );
 46         //for( int i=1;i<=n;i++ ){
 47         //    printf("id:%d time:%d sum:%d\n",a[i].id,a[i].time,a[i].sum);
 48         //}
 49     //    printf("\n");
 50         int f4=1,f3=1,f2=1,f1=1;
 51         //f1=f2=f3=f4=0;
 52         /*
 53         if(n1==1)n1++;
 54         if(n2==1)n2++;
 55         if(n3==1)n3++;
 56         if(n4==1)n4++;
 57         */
 58         for( int i=1;i<=n;i++ ){
 59             if( a[i].sum==5 ){
 60                  a[i].ans=100;//printf("100\n");
 61                  continue;
 62             }
 63             
 64             if( a[i].sum==0 ){
 65                 a[i].ans=50;//printf("50\n");
 66                 continue;
 67             }
 68             
 69             if( a[i].sum==4 && f4<=(n4/2) ){
 70                 f4++;
 71                 a[i].ans=95;//printf("95\n");
 72                 continue;
 73             }
 74             else if( a[i].sum==4 ){
 75                 f4++;
 76                 a[i].ans=90;//printf("90\n");
 77                 continue;
 78             }
 79             
 80             if( a[i].sum==3 && f3<=(n3/2) ){
 81                 f3++;
 82                 a[i].ans=85;//printf("85\n");
 83                 continue;
 84             }
 85             else if( a[i].sum==3 ){
 86                 f3++;
 87                 a[i].ans=80;//printf("80\n");
 88                 continue;
 89             }
 90             
 91             if( a[i].sum==2 && f2<=(n2/2) ){
 92                 f2++;
 93                 a[i].ans=75;//printf("75\n");
 94                 continue;
 95             }
 96             else if( a[i].sum==2 ){
 97                 f2++;
 98                 a[i].ans=70;//printf("70\n");
 99                 continue;
100             }
101             
102             if( a[i].sum==1 && f1<=(n1/2) ){
103                 f1++;
104                 a[i].ans=65;//printf("65\n");
105                 continue;
106             }
107             else if( a[i].sum==1 ){
108                 f1++;
109                 a[i].ans=60;//printf("60\n");
110                 continue;
111             }
112             
113         }
114         sort( a+1,a+1+n,cmp2 );
115         for( int i=1;i<=n;i++ ){
116         //    printf("id:%d time:%d ",a[i].id,a[i].time);
117             printf("%d\n",a[i].ans);
118         }
119         printf("\n");
120     }
121     return 0;
122 }
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