#include <stdio.h>
int n;
double a;
double t;
int g;
int open_or_close[2000000] = { 0 };
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%lf", &a);
scanf("%lf", &t);
for (int k = 1; k <= t; k++)
{
double m = (double)k;
g = m * a;
open_or_close[g] = (open_or_close[g] + 1) % 2;
}
}
//小数相乘后取整数的部分
//float强制转换为int就能实现小数部分的截断
for (int u = 0; u < 2000000; u++)
{
if (open_or_close[u] == 1)
{
printf("%d", u);
break;
}
}
}
因篇幅问题不能全部显示,请点此查看更多更全内容
Copyright © 2019- bangwoyixia.com 版权所有 湘ICP备2023022004号-2
违法及侵权请联系:TEL:199 1889 7713 E-MAIL:2724546146@qq.com
本站由北京市万商天勤律师事务所王兴未律师提供法律服务