codeforces Round #267(div2) C解題報告

C. George and Job time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.

Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers:

[l1, r1], [l2, r2], ..., [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < ... < lk ≤ rk ≤ n; ri - li + 1 = m), 

in such a way that the value of sum 

 is maximal possible. Help George to cope with the task.

Input

The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn(0 ≤ pi ≤ 109).

Output

Print an integer in a single line — the maximum possible value of sum.

Sample test(s) input

5 2 1
1 2 3 4 5
      

output

9
      

input

7 1 3
2 10 7 18 5 33 0
      

output

61      

題目大意：

給出一個數字序列，要求找出k個m長度不相交的區間，且區間數字之和最大

解法：

經典的動态規劃。狀态轉移方程：f[i][j] = max(f[i][j-1], f[i-1][j-m]+sum[j]-sum[j-m]) 其中i代表有多少段(1 <= i <= k)，j表示第幾個數字。

注意要使用long long，可以用滾動數組優化記憶體

代碼：

#include <cstdio>

#define LL long long

LL n, m, k, f[2][5010], sum[5010];

LL max(LL a, LL b) {
	if (a > b) return a;
	return b;
}

void init() {
	scanf("%I64d%I64d%I64d", &n, &m, &k);
	for (int i = 1; i <= n; i++) {
		scanf("%I64d", &sum[i]);
		sum[i] += sum[i-1];
	}
}

void solve() {
	LL ans=0;

	for (int i = 1; i <= k; i++) 
		for (int j = m; j <= n; j++)
			f[i&1][j] = max(f[i&1][j-1] ,f[(i-1)&1][j-m]+sum[j]-sum[j-m]);
				
	for (int j = 1; j <= n; j++)
		ans = max(ans, f[k&1][j]);
	printf("%I64d\n", ans);
}

int main() {
	init();
	solve();
}