1.
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% 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
#include<stdio.h>
#include<algorithm>
using namespace std;
struct goods {
double j; //物品总重
double f; //物品总价值
double s; //物品性价比
bool operator < (const goods &A) const {
return s > A.s;
}
}buf[1000];
int main()
{
double m;
int n;
while (scanf("%lf%d", &m, &n) != EOF) {
if (m == -1 && n == -1)
break;
for (int i = 0; i < n; i++) {
scanf("%lf%lf", &buf[i].j, &buf[i].f);
buf[i].s = buf[i].j / buf[i].f; //计算性价比
}
sort(buf, buf + n);
int idx = 0; //当前货物下标
double ans = 0;//累加所能得到的总重量
while (m > 0 && idx < n) {
//物品剩余（idx<n）且钱剩余（m>0）时继续循环
if (m > buf[idx].f) { //若能买下全部物品
ans += buf[idx].f;
m -= buf[idx].f;
}
else { //若只能买下部分该物品
ans += buf[idx].j*m / buf[idx].f;
m = 0;
}
idx++;
}
printf("%.3lf\n", ans);
}
return 0;
}Problem Description
“今年暑假不AC？”
“是的。”
“那你干什么呢？”
“看世界杯呀，笨蛋！”
“@#$%^&*%...”
确实如此，世界杯来了，球迷的节日也来了，估计很多ACMer也会抛开电脑，奔向电视了。
作为球迷，一定想看尽量多的完整的比赛，当然，作为新时代的好青年，你一定还会看一些其它的节目，比如新闻联播（永远不要忘记关心国家大事）、非常6+7、超级女生，以及王小丫的《开心辞典》等等，假设你已经知道了所有你喜欢看的电视节目的转播时间表，你会合理安排吗？（目标是能看尽量多的完整节目）
Input
输入数据包含多个测试实例，每个测试实例的第一行只有一个整数n(n<=100)，表示你喜欢看的节目的总数，然后是n行数据，每行包括两个数据Ti_s,Ti_e (1<=i<=n)，分别表示第i个节目的开始和结束时间，为了简化问题，每个时间都用一个正整数表示。n=0表示输入结束，不做处理。
Output
对于每个测试实例，输出能完整看到的电视节目的个数，每个测试实例的输出占一行。
Sample Input
12 1 3 3 4 0 7 3 8 15 19 15 20 10 15 8 18 6 12 5 10 4 14 2 9 0
Sample Output
5
#include<iostream>
#include<algorithm>
using namespace std;
struct program {
int starttime;
int endtime;
bool operator < (const program &A) const {
return endtime<A.endtime;
}
}buf[100];
int main()
{
int n;
while (scanf("%d", &n) != EOF) {
if (n == 0)
break;
for (int i = 0; i < n; i++) {
scanf("%d%d", &buf[i].starttime, &buf[i].endtime);
}
sort(buf, buf + n);
int currenttime = 0, ans = 0;
for (int i = 0; i < n; i++) {
if (currenttime <= buf[i].starttime) {
currenttime = buf[i].endtime;
ans++;
}
}
printf("%d\n", ans);
}
return 0;
}