UVALive - 2678 二分/尺取

题意:求最小的长度L满足该长度上的元素和大于等于S

最近dp做多了总有一种能用dp解决一切的错觉

二分长度解决

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e5+15;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
int n,s;
ll sum[maxn],a[maxn];
bool C(int len){
    for(int i=1;i+len-1<=n;i++){
        if(sum[i+len-1]-sum[i-1]>=s)return 1;
    }
    return 0;
}
int main(){
    while(cin>>n>>s){
        rep(i,1,n)scanf("%lld",&a[i]);
        rep(i,1,n) sum[i]=sum[i-1]+a[i];
        int l=1,r=n,ans;
        while(l<r){
            int mid=l+(r-l)/2;
            if(C(mid))r=mid;
            else l=mid+1;
        }
        println((C(l)?l:0));
    }
    return 0;
} 

原文地址：https://www.cnblogs.com/caturra/p/8726785.html