Problem Description
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a%
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a%
pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning
this homework to you: tell him the maximum amount of JavaBeans he can obtain.
Input
The input consists of multiple test cases. Each test case begins with a line containing two non-negative
integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.
Output
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the
maximum amount of JavaBeans that FatMouse can obtain.
Sample Input
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
Sample Output
13.333
31.500
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#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include <math.h> #define SIZE 1001 int J[SIZE]; int F[SIZE]; double rate[SIZE]; int bigger(int a, double keyRate, double keyJ); int main() { int M, N; int i, j, tempIndex; double tempRate,tempJ,tempF; double total; while (scanf("%d%d", &M, &N) == 2) { total = 0; if (M == -1 && N == -1) break; for (i = 0; i < N; i++) { scanf("%d%d", &J[i], &F[i]); rate[i] = ((double)F[i] / J[i]); } //插入排序 for (i = 1; i < N; i++) { tempRate = rate[i]; tempJ = J[i]; tempF = F[i]; tempIndex = i; j = i - 1; while (j >= 0 && bigger(j, tempRate, tempJ)) { rate[j + 1] = rate[j]; J[j + 1] = J[j]; F[j + 1] = F[j]; j = j - 1; } rate[j + 1] = tempRate; J[j + 1] = tempJ; F[j + 1] = tempF; } for (i = 0; i < N; i++) { //判断比率为0,也就是免费送的情况 if (M > F[i] || rate[i] <= 0.0000001) { total += J[i]; M -= F[i]; } else { total += (double)M / F[i] * J[i]; break; } } printf("%0.3f\n", total); } } int bigger(int a, double keyRate, double keyJ) { if (fabs(rate[a] - keyRate) <= 0.000001) return J[a] < keyJ; else { return rate[a] > keyRate; } }