kotori的设备--二分答案

Luogu 3743

题目分析：

• 如果机器每秒总的消耗$<=p$，则输出$-1$
• 二分时间，注意$r=10^{10},eps=10^{-6}$
• 求出充电宝能充的总能量$pdc$，对于一个机器能量不够就给它充，最后判断$pdc$是否小于$0$

Code：

#include <bits/stdc++.h>
using namespace std;
#define eps 1e-6
#define maxn 100010int n;
double sum=0,m,a[maxn],b[maxn];inline int read_() {int x=0,f=1;char c=getchar();while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}return x*f;
}inline bool check_(double x) {double pdc=x*m;double AKIOI;for(int i=1;i<=n;++i) {if(pdc<0) return false;AKIOI=a[i]*x;if(AKIOI<=b[i]) continue;else {AKIOI-=b[i];pdc-=AKIOI;}}if(pdc>=0) return true;else return false;
}void readda_() {n=read_();scanf("%lf",&m);for(int i=1;i<=n;++i) {scanf("%lf%lf",&a[i],&b[i]);sum+=a[i];}if(m>=sum) printf("-1");else {double l=0,ans,r=10000000000,mid;while(r-l>eps) {mid=(l+r)/2;if(check_(mid)) {l=mid;ans=mid;}else r=mid;} printf("%.10lf",ans);}
}int main() {freopen("a.txt","r",stdin);readda_();return 0;
}