标签:des style java color os 2014
<span style="color:#330099;">/* K - 贪心 基础 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status 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% 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 BY Grant Yuan 2014.7.14 贪心 */ /* 描述 有一个投资人,他有金钱m,可选投资项目n个。对于每个投资项目,投入金钱c则获得收益f。要求对每个项目所投入的资金r应该在0到c之间(即0<=r<=c),获得的收益为r*f/c。现在你的任务就是求出投资人能获得的最大收益。 输入 包括多个测例,每个测例第一行为两个整数m,n,分别表示金钱数和项目个数。接下来n行,每行两个整数为收益f和投入c。输入最后以两个-1结尾。 输出 包括n行,每行为对应测例的最大收益,保留三位小数printf("%.3lf\n",result);。 输入样例 3 3 4 2 2 1 3 1 4 3 4 2 5 3 4 3 -1 -1 输出样例 7.000 7.333 By yuan.c 20146/22 提示 */ #include<stdio.h> #include<iostream> #include<cstring> #include<cstdlib> using namespace std; int N; int M; double p[1002]; int m[1002]; int s[1002]; double sum=0; void sort() {int t;double l; for(int i=0;i<N-1;i++) for(int j=i;j<N;j++) { if(p[i]<p[j]){ l=p[i],p[i]=p[j],p[j]=l; t=m[i],m[i]=m[j],m[j]=t; t=s[i],s[i]=s[j],s[j]=t; } } } int main() { while(1){ scanf("%d %d",&M,&N); if(N==-1&&M==-1) break; for(int i=0;i<N;i++) scanf("%d %d",&s[i],&m[i]); for(int i=0;i<N;i++) p[i]=s[i]*1.0/m[i]*1.0; sort(); for(int i=0;i<N;i++) if(M>=m[i]) sum+=s[i],M-=m[i]; else {sum+=M*p[i]; break; } printf("%.3lf\n",sum); sum=0; memset(s,0,100); memset(m,0,100); memset(p,0,100); } return 0; } </span>
标签:des style java color os 2014
原文地址:http://blog.csdn.net/yuanchang_best/article/details/37774377