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;}