uva 10670 Work Reduction

题意：有N个任务，求在完成的只剩M个的时候花费最少的方法，坑的地方是rounding down是向下取整的意思，至于方法的选择只要采用贪心就行了，对比一下两种花费就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 105;

struct node
{
    char name[30];
    int A,B;
    int cost;
}arr[MAXN];
int N,M,L;

bool cmp(node a,node b)
{
    if (a.cost != b.cost)
        return a.cost < b.cost;
    return strcmp(a.name,b.name) < 0;
}

int main()
{
    int t,cas = 1;
    scanf("%d",&t);
    while (t--)
    {
        char str[30];
        scanf("%d%d%d",&N,&M,&L);
        for (int i = 0; i < L; i++)
        {
            scanf("%s",str);
            int j;
            for (j = 0; str[j] != ':'; j++)
                arr[i].name[j] = str[j];
            arr[i].name[j] = '\0';
            sscanf(str+j+1,"%d,%d",&arr[i].A,&arr[i].B);
            arr[i].cost = 0;
        }

        for (int i = 0; i < L; i++)
        {
            int cur = N;
            int A = arr[i].A,B = arr[i].B;
            int half = (cur + 1) / 2;
            while (cur - half >= M && B <= half * A)
            {
                arr[i].cost += B;
                cur -= half;
                half = (cur + 1) / 2;
            }
            if (cur > M)
                arr[i].cost += (cur - M) * A;
        }
        
        sort(arr,arr+L,cmp);
        printf("Case %d\n",cas++);
        for (int i = 0; i < L; i++)
            printf("%s %d\n",arr[i].name,arr[i].cost);
    }
    return 0;
}