题目大意：对n份文件进行处理使其减少到m份，有l个机构可供选择。每个机构提供两种方案：每减少一份收费a元，或者减少到文件数量的一半收费b元。根据各个机构收取费用进行排序。
很直接的题目，直接进行模拟就好了。其中对A、B两种方案的选择使用贪心策略。
1 #include <cstdio> 2 #include <cctype> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 #define MAXN 100+10 7 8 struct Agency 9 { 10 char name[20]; 11 int cost; 12 }; 13 Agency agency[MAXN]; 14 15 bool cmp (const Agency& a, const Agency& b) 16 { 17 if (a.cost != b.cost) return a.cost < b.cost; 18 else return strcmp(a.name, b.name) < 0; 19 } 20 21 int main() 22 { 23 #ifdef LOCAL 24 freopen("in", "r", stdin); 25 //freopen("out", "w", stdout); 26 #endif 27 int T; 28 scanf("%d", &T); 29 for (int kase = 1; kase <= T; kase++) 30 { 31 int n, m, l; 32 scanf("%d%d%d", &n, &m, &l); 33 getchar(); 34 for (int i = 0; i < l; i++) 35 { 36 int cnt = 0; 37 char ch = getchar(); 38 while (isupper(ch)) 39 { 40 agency[i].name[cnt++] = ch; 41 ch = getchar(); 42 } 43 agency[i].name[cnt] = ' '; 44 int a, b; 45 scanf("%d,%d", &a, &b); 46 getchar(); 47 agency[i].cost = 0; 48 int remain = n; 49 while (remain > m) 50 { 51 int half = remain / 2; 52 if (half < m) 53 { 54 agency[i].cost += (remain - m) * a; 55 remain = m; 56 } 57 else 58 { 59 int t = a * (remain - half); // the cost reducing from remain to half with A 60 if (t < b) agency[i].cost += t; 61 else agency[i].cost += b; 62 remain = half; 63 } 64 } 65 } 66 sort(agency, agency+l, cmp); 67 printf("Case %d ", kase); 68 for (int i = 0; i < l; i++) 69 printf("%s %d ", agency[i].name, agency[i].cost); 70 } 71 return 0; 72 }
不过在读取名字的while循环中如果使用 while (ch = getchar() && isupper(ch)) 就会出问题，不知道为什么...