尺取法 POJ 3601 Subsequence

题目传送门

 /*
     题意：求连续子序列的和不小于s的长度的最小值
     尺取法：对数组保存一组下标(起点，终点)，使用两端点得到答案
         1. 记录前i项的总和，求[i, p)长度的最小值，用二分找到sum[p] - s[i] >= s的p
         2. 除了O (nlogn)的方法，还可以在O (n)实现，[i, j)的区间求和，移动两端点，更新最小值，真的像尺取虫在爬:)
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;

 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN];
 ll sum[MAXN];

 int main(void)        //POJ 3601 Subsequence
 {
     int t;    scanf ("%d", &t);
     while (t--)
     {
         memset (sum, , sizeof (sum));
         int n, s;
         scanf ("%d%d", &n, &s);
         for (int i=; i<=n; ++i)    {scanf ("%d", &a[i]);    sum[i] = sum[i-] + a[i];}

         if (sum[n] < s)    {puts ("");    continue;}

         int ans = n;
         for (int i=; sum[i]+s<=sum[n]; ++i)
         {
             int p = lower_bound (sum+i, sum++n, sum[i] + s) - sum;
             ans = min (ans, p - i);
         }

         printf ("%d\n", ans);
     }

     return ;
 }

 /*
 2
 10 15
 5 1 3 5 10 7 4 9 2 8
 5 11
 1 2 3 4 5
 */

 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;

 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN];

 int main(void)        //POJ 3601 Subsequence
 {
     int t;    scanf ("%d", &t);
     while (t--)
     {
         int n, s;
         scanf ("%d%d", &n, &s);
         for (int i=; i<=n; ++i)    scanf ("%d", &a[i]);

         int ans = n + ;    int i = , j = ;    ll sum = ;
         while ()
         {
             while (j <= n && sum < s)    sum += a[j++];
             if (sum < s)    break;
             ans = min (ans, j - i);
             sum -= a[i++];
         }

         if (ans == n + )    puts ("");
         else    printf ("%d\n", ans);
     }

     return ;
 }

 /*
 2
 10 15
 5 1 3 5 10 7 4 9 2 8
 5 11
 1 2 3 4 5
 */

O (n)