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 }