Description
Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.
Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S 1, S 2, S 3 ... S n.
Process to the end of file.
Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
Sample Output
6
8
Hint
Huge input, scanf and dynamic programming is recommended.
/* 题意：给你一个长度为n的序列，让你求出最大m个字段的序列元素和 初步思路：动态规划最大m字段和,dp数组，dp[i][j]表示以a[j]结尾的，i个字段的最大和 两种情况：1.第a[j]元素单独作为第i个字段 2.第a[j]元素和前面的字段共同当做第i个字段 得到状态转移方程：dp[i][j]=max( dp[i][j-1]+a[j] , max(dp[i-1][t])+a[j]); 但是实际情况是，时间复杂度和空间复杂度都是相当的高，所以要进行时间和空间的优化： 将每次遍历的时候的max(dp[i-1][t]) 用一个数组d储存起来，这样就能省去寻找max(dp[i-1][t])的时间， 这样状态转移方程就变成了 dp[i][j]=max( dp[i][j-1]+a[j] , d[j-1]+a[j]), 会发现dp数组的可以 省去一维，因为每次都是和前一次的状态有关，所以可以记录前一次状态,再用一个变量tmp记录下dp[i][j-1]， 这样方程就变成了 dp[i][j]=max( tmp+a[j] , d[j-1]+a[j])；这样就可以化简一下就是：dp[i][j]= max( tmp , d[j-1])+a[j]; */ #include <bits/stdc++.h> #define N 1000005 using namespace std; int a[N]; int n,m; int d[N];//用来存储j-1的位置用来存储 max(dp[i-1][t]) int main(){ // freopen("in.txt","r",stdin); while(scanf("%d%d",&m,&n)!=EOF){ memset(d,0,sizeof d); for(int i=1;i<=n;i++){ scanf("%d",&a[i]); } /* dp[i][j]=max( dp[i][j-1]+a[j] , max(dp[i-1][t])+a[j]) */ for(int i=1;i<=m;i++){//遍历字段 int tmp = 0;//用来记录dp[i-1][j] for(int k = 1; k <= i; ++k) tmp += a[k]; //由于d[n]的位置是永远都用不到的，所以就用来存储最后的姐 d[n] = tmp;//前面的i项，每项都是一个段的时候 for(int j = i+1; j <= n; ++j) { tmp = max(d[j-1], tmp) + a[j]; //a[j]单独作为一个段的情况 和 前面的max(dp[i-1][t]) d[j-1] = d[n];//将这个值保存下来 d[n] = max(d[n], tmp); //比较大小方便答案的输出 } } printf("%d ",d[n]); } return 0; }