uva10670 - Work Reduction

题意：
老板给你n个任务，你只能最多留m个，你干不了就辞了你，现在一个机构能帮你完成任务，有A,B两种方案，金额不同，A能帮你完成1个，B能帮你完成当前的一半。
思路：
贪心的思路，题目要求的其实就是关于A，B的选择，如过当前扣掉一半后所得>=m而且经费画得比A少就选B，否则就选择A
代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 105;struct comp{    char name[20];    int A, B;    int cnt;}a[N];char str[N];int cmp(comp a,comp b) {    if (a.cnt != b.cnt)        return a.cnt < b.cnt;    else        return strcmp(a.name, b.name) < 0;}int n, m, l;int main() {    int cas, Case = 0;    scanf("%d", &cas);    while (cas--) {        scanf("%d%d%d\n", &n, &m, &l);        for (int i = 0; i < l; i++) {            //scanf_s("%s:%d,%d\n", a[i].name, &a[i].A, &a[i].B);            gets(str);            int j;            for (j = 0; str[j] != ':'; ++j)                a[i].name[j] = str[j];            a[i].name[j] = '\0';            sscanf(str + j + 1, "%d,%d", &a[i].A, &a[i].B);            a[i].cnt = 0;        }        for (int i = 0; i < l; i++) {            int left = n;            int A = a[i].A, B = a[i].B;            int mid = (left + 1) / 2;            while (left - mid >= m &&B <= mid*A) {                a[i].cnt += B;                left -= mid;                mid = (left + 1) / 2;            }            if (left>m)                a[i].cnt += (left - m)*A;        }        printf("Case %d\n", ++Case);        sort(a, a + l, cmp);        for (int i = 0; i < l; i++)            printf("%s %d\n",a[i].name,a[i].cnt);    }    return 0;}