cf176B. Pipeline

//题目连接:<a target=_blank href="http://codeforces.com/contest/287/problem/B">点击打开链接</a>

#include<iostream>
#include<cstdio>
using namespace std;
inline long long sum(long long a,long long k){
    return k*(k+1)/2-(k-a)*(k-a+1)/2-a+1;
}
int f(long long n,long long k){
    long long i,tot=1+(k-1)*k/2;
    if(n>tot) return -1;
    int toright=0;
    long long l=0,r=k,mid=(l+r)/2;//the numbers,have to take into account that l+1=r;uncontinue
    while(l<=r){
        long long tmp=sum(mid,k);
        if(tmp==n)break;
        else if(tmp<n){
            if(sum(mid+1,k)>=n){//对于mid个排水口,有2,3,...,mid+1的min,也有k,k-1,k-2,k-mid+1的max,而我的sum表示的是max,//sum(mid)和sum(mid+1)之间不是连续的,所以加上if判断,修正
                toright=1;
                break;
            }
            l=mid+1;
            mid=(l+r)/2;
        }
        else{
            if(sum(mid-1,k)<n) break;
            r=mid-1;
            mid=(l+r)/2;
        }
    }
    return mid+toright;
}

int main(){
    long long n,k;
    scanf("%I64d%I64d",&n,&k);
    cout<<f(n,k)<<endl;
    return 0;
}

原文地址：https://www.jb51.cc/javaschema/285423.html