LA 2678 Subsequence

题意：给出n个整数，给定整数S,求长度最短的连续序列，使它们的和大于或等于S

第一种，是求前缀和，然后用lower_bound来找

第二种，尺取法，

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=100005;

int n,S;
int a[maxn],sum[maxn];

int main(){
    while(scanf("%d %d",&n,&S) != EOF){
        memset(a,0,sizeof(a));
        memset(sum,0,sizeof(sum));
        for(int i = 0;i < n;i++) scanf("%d",&a[i]);
        sum[0] = a[0];
        for(int i = 1;i < n;i++) sum[i] = sum[i-1] + a[i];
        
    //    for(int i =0;i <= n;i++)
    //    printf("sum[%d] = %d
",i,sum[i]);
        if(sum[n-1] < S){
            printf("0
");
        }
        else{
            int res = INF;
        for(int s = 0;sum[s] + S <= sum[n-1];s++){
            int t = lower_bound(sum,sum + n,sum[s] + S) - sum;
            res = min(res,t-s);
        //    printf("res = %d
",res);
        }
        printf("%d
",res);
        }        
    }
    return 0;
}

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=100005;

int a[maxn];
int n,S;

void solve(){
    int t = 0,s = 0;
    int res = INF;
    int sum = 0;
    for(;;){
        while(t < n && sum < S){
            sum += a[t++];
        }
        if(sum < S) break;
        res = min(res,t - s);
        sum -= a[s++];
    }
    if(res > n) res = 0;
    printf("%d
",res);
}

int main(){
    while(scanf("%d %d",&n,&S) != EOF){
        memset(a,0,sizeof(a));
        for(int i = 0;i < n;i++) scanf("%d",&a[i]);
        solve();
    }
    return 0;
}