题目链接：https://vjudge.net/problem/FZU-2013
Problem 2013 A short problemAccept: 356 Submit: 1083
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
The description of this problem is very short. Now give you a string(length N), and ask you the max sum of the substring which the length can't small than M.
Input
The first line is one integer T(T≤20) indicates the number of the test cases. Then for every case, the first line is two integer N(1≤N≤1000000) and M(1≤M≤N).
Then one line contains N integer indicate the number. All the number is between -10000 and 10000.
Output
Output one line with an integer.
Sample Input
2
5 1
1 -2 -2 -2 1
5 2
1 -2 -2 -2 1
Sample Output
1
-1
Source
FOJ有奖月赛-2011年03月题意：
给出一个序列，求长度不小于m且和最大的子序列。
题解：
1.求出前缀和。
2.将0插入到线段树第0位，然后从第m为开始枚举：去线段树范围为0~i-m的地方找最小值，然后以i为结尾的子序列最大值即为：sum[i] - query(1,0,n,0,i-m)。答案即为max（sum[i] - query(1,0,n,0,i-m)）。
3.类似的题目：CSU - 1551 Longest Increasing Subsequence Again
线段树：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const int INF = 2e9; 15 const LL LNF = 9e18; 16 const int MOD = 1e9+7; 17 const int MAXN = 1e6; 18 19 int minv[MAXN*4]; 20 void add(int u, int l, int r, int x, int val) 21 { 22 if(l==r) 23 { 24 minv[u] = val; 25 return; 26 } 27 int mid = (l+r)/2; 28 if(x<=mid) add(u*2, l, mid, x, val); 29 else add(u*2+1, mid+1,r, x, val); 30 minv[u] = min(minv[u*2], minv[u*2+1]); 31 } 32 33 int query(int u, int l, int r, int x, int y) 34 { 35 if(l<=x&&y<=r) 36 return minv[u]; 37 38 int mid = (l+r)/2; 39 int ret = INF; 40 if(x<=mid) ret = min(ret, query(u*2, l, mid, x, mid)); 41 if(y>=mid+1) ret = min(ret, query(u*2+1, mid+1, r, mid+1, y)); 42 return ret; 43 } 44 45 int sum[MAXN]; 46 int main() 47 { 48 int T, n, m; 49 scanf("%d", &T); 50 while(T--) 51 { 52 scanf("%d%d", &n,&m); 53 sum[0] = 0; 54 for(int i = 1; i<=n; i++) 55 { 56 int val; 57 scanf("%d", &val); 58 sum[i] = sum[i-1]+val; 59 } 60 61 for(int i = 0; i<=n*4; i++) 62 minv[i] = INF; 63 64 int ans = -INF; 65 add(1,0,n,0,0); 66 for(int i = m; i<=n; i++) 67 { 68 ans = max(ans, sum[i]-query(1,0,n,0,i-m)); 69 add(1,0,n,i-m+1,sum[i-m+1]); 70 } 71 printf("%d ", ans); 72 } 73 }
树状数组：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const int INF = 2e9; 15 const LL LNF = 9e18; 16 const int MOD = 1e9+7; 17 const int MAXN = 1e6; 18 19 int T, n, m; 20 int lowbit(int x) 21 { 22 return x&(-x); 23 } 24 25 int c[MAXN]; 26 void add(int x, int d) 27 { 28 while(x<=n) 29 { 30 c[x] = min(c[x], d); 31 x += lowbit(x); 32 } 33 } 34 35 int query(int x) 36 { 37 int s = INF; 38 while(x>0) 39 { 40 s = min(c[x], s); 41 x -= lowbit(x); 42 } 43 return s; 44 } 45 46 int sum[MAXN]; 47 int main() 48 { 49 scanf("%d", &T); 50 while(T--) 51 { 52 memset(c, 0, sizeof(c)); 53 scanf("%d%d", &n,&m); 54 sum[0] = 0; 55 for(int i = 1; i<=n; i++) 56 { 57 int val; 58 scanf("%d", &val); 59 sum[i] = sum[i-1]+val; 60 } 61 for(int i = 1; i<=n+1; i++) 62 c[i] = INF; 63 64 int ans = -INF; 65 add(1, 0); 66 for(int i = m; i<=n; i++) 67 { 68 ans = max(ans, sum[i]-query(i-m+1)); 69 add(i-m+2, sum[i-m+1]); 70 } 71 printf("%d ", ans); 72 } 73 }