Ac_dream 1224 robbers (Greedy)

The money that the n robbers snatched was K1, K2, K3,.... The total money was M,
However, in the case of money sharding, it is allocated according to the ratio of x1/y, x2/y, X3/y! In this case
Some robbers may feel unfair. The unfair level is | XI/Y-ki/M | (floating point operation), and a sequence Ki is output to make
The minimum injustice .....

Idea: greedy! First, calculate the money (NI) that each person receives according to the [XI/y] (rounded up) ratio, and then get
The remaining amount of money is found to be less than the relative proportion allocated to everyone, that is, the maximum value of XI/y * m-ni! Find this person
Then, increase the amount of money he gets to 1!

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #define N 1006 5 using namespace std; 6  7 int num[N]; 8 int x[N]; 9 bool flag[N];10 11 int main(){12     int n, m, y;13     while(scanf("%d%d%d", &n, &m, &y) != EOF){14         memset(flag, 0, sizeof(flag));15         int left = 0;16         for(int i=1; i<=n; ++i){17             scanf("%d", &x[i]);18             num[i] = x[i]*m/y;19             left += num[i];20             if( x[i]%y == 0 )  flag[i] = true;21         }22          left = m - left;23         while(left > 0){24             double tmp = 0.0;25             int p = 0;26             for(int i=1; i<=n; ++i)27                 if(tmp < x[i]*1.0/y * m - num[i]){28                     tmp = x[i]*1.0/y * m - num[i];29                     p = i;30                 } 31             --left;32             ++num[p];33         }34         printf("%d", num[1]);35         for(int i=2; i<=n; ++i)36             printf(" %d", num[i]);37         printf("\n");38     }39     return 0;40 } 

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Ac_dream 1224 robbers (Greedy)