Kattis speed 二分答案

题目大意:速度表有误差C,速度表测的速度是S,而真正的速度是S + C。现在走了n(1<=n<=1000)段路程,总时间是t(1<=t<=1e6),每段路程的距离是di, 测得的速度是si,求误差C。

二分答案:∑di/(C+si)=t 二分C即可

#include <cstdio>

using namespace std;
const int N = 1010;
const double esp = 1e-6;
int dis[N], read[N];
int n;
double t;

bool solve(double mid)
{
    double sumt = 0.0;
    for(int i = 1; i <= n; ++i)
    {
        if(mid+read[i]<=0) return 0;
        sumt += 1.0*(dis[i]/(read[i]+mid));
    }
    return sumt < t;
}

double bin(double l, double r)
{
    double mid = (l+r)/2;
    while(r-l > esp)
    {
        if(solve(mid)) r = mid;
        else l = mid;
        mid = (l+r)/2;//mid放前面不行就放后面
    }
    return mid;
}

int main()
{
    scanf("%d %lf", &n, &t);
    for(int i = 1; i <= n; ++i)
        scanf("%d %d", &dis[i], &read[i]);
    double l = -1e9;
    double r = 1e9;
    double ans = bin(l, r);
    printf("%.6f\n", ans);
    return 0;
}

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转载自blog.csdn.net/jay__bryant/article/details/81192034
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