POJ3061 Subsequence 尺取or二分

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <string.h>
 5 using namespace std;
 6 
 7 const int INF = 2147483647;
 8 int a[100010];
 9 int main()
10 {
11     int S, t, n;
12     int T, mi;
13     int l, r, mid;
14     while(cin >> T)
15     {
16         while(T--)
17         {
18             cin >> n >> S;
19             for(int i = 1; i <= n; i++)
20             {
21                 scanf("%d", &t);
22                 if(i == 1)
23                     a[i] = t;
24                 else
25                     a[i] = a[i-1] + t;
26             }
27 
28             a[0] = 0;
29             mi = INF;
30             for(int i = 1; i <= n; i++)
31             {
32                 l = i;
33                 r = n;
34                 while(l <= r)
35                 {
36                     mid = (l + r)/2;
37                     if( a[mid] - a[i-1] >= S)
38                     {
39                         if( mi > mid-i+1 )
40                             mi = mid - i + 1;
41                         r = mid - 1;
42                     }
43                     else
44                     {
45                         l = mid + 1;
46                     }
47                 }
48             }
49             if(mi == INF)
50                 printf("0\n");
51             else printf("%d\n", mi);
52 
53 
54 
55         }
56     }
57     return 0;
58 }

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <string.h>
 5 using namespace std;
 6 
 7 int a[100010];
 8 const int INF = 2147483647;
 9 int main()
10 {
11     int T, n, S, sum;
12     int l, r, mi;
13     
14     while(cin >> T)
15     {
16         while(T--)
17         {
18             cin >> n >> S;
19             for(int i = 0; i < n; i++)
20             {
21                 scanf("%d", a + i);
22             }
23             
24             l = r = sum = 0;
25             mi = INF;
26             for(;;)
27             {
28                 while(r < n && sum  < S)
29                 {
30                     sum += a[r++];
31                 }
32                 if( sum < S)
33                     break;
34                 else 
35                 {
36                     mi = min(mi, r - l);
37                     sum -= a[l++];
38                 }
39             }
40             
41             
42             if(mi == INF)
43                 printf("0\n");
44             else printf("%d\n", mi);
45         }
46     }
47     return 0;
48 }