题目大意:
给一个序列,要求将序列分成m段,从左至右每一段分别长l1,l2,...lm,求最大的和是多少。
题目分析:
和最大m段子段和相似,先枚举(i in [1,m]),然后$j in [num[m], n] $,dp转移为: $$dp[j][i] = max(dp[j - 1][i], dp[j - num[i]][i - 1] + sum[j] - sum[i - num[i]])$$
code:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
//#include<vector>
using namespace std;
const int N = 1005, M = 25;
int n, m, num[N];
typedef long long ll;
ll sum[N], f[N][N];
inline int read(){
ll i = 0, f = 1; char ch = getchar();
for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
if(ch == '-') f = -1, ch = getchar();
for(; ch >= '0' && ch <= '9'; ch = getchar())
i = (i << 3) + (i << 1) + (ch - '0');
return i * f;
}
inline void wr(ll x){
if(x < 0) putchar('-'), x = -x;
if(x > 9) wr(x / 10);
putchar(x % 10 + '0');
}
int main(){
while(n = read()){
m = read();
sum[0] = 0;
for(int i = 1; i <= m; i++) num[i] = read();
for(int i = 1; i <= n; i++){
ll x = read() * 1LL;
sum[i] = sum[i - 1] + x;
}
int now = 0;
for(int i = 1; i <= m; i++){
now += num[i];
for(int j = now; j <= n; j++)
f[j][i] = max(f[j - 1][i], f[j - num[i]][i - 1] + sum[j] - sum[j - num[i]]);
}
wr(f[n][m]), putchar('
');
}
}