HDU 1244 Max sum plus

Although this question looks like HDU 1024 Max sum plus, it seems easier than 1024.

Previously wa several times, because I started to write DP [22] [maxn] as DP [maxn] [22], orz

It seems that arrays out of bounds may not necessarily cause program crash or return an error.

 

DP [I] [J] indicates the maximum number of J numbers that constitute the first I segment.

State transition equation:

DP [I] [J] = max {DP [I] [J-1], DP [I-1] [J-len [I] + sum [J]-sum [J-len [I]}

Corresponding to not take the I number and take the J number and the adjacent total Len [I] Number respectively as the I segment plus the previous number constitute the maximum value of the I-1 segment

 

 1 //#define LOCAL 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6  7 const int maxn = 1000 + 10; 8 int a[maxn], sum[maxn], len[22], dp[22][maxn]; 9 10 int main(void)11 {12     #ifdef LOCAL13         freopen("1244in.txt", "r", stdin);14     #endif15 16     int n, m;17     while(scanf("%d", &n) == 1 && n)18     {19         scanf("%d", &m);20         for(int i = 1; i <= m; ++i)21             scanf("%d", &len[i]);22         for(int i = 1; i <= n; ++i)23             scanf("%d", &a[i]);24         sum[0] = 0;25         for(int i = 1; i <= n; ++i)26             sum[i] = sum[i - 1] + a[i];27         memset(dp, 0, sizeof(dp));28         for(int i  = 1; i <= m; ++i)29         {30             for(int j = 1; j < len[i]; ++j)31                 dp[i][j] = dp[i][j-1];32             for(int j = len[i]; j <= n; ++j)33                 dp[i][j] = max(dp[i][j-1], dp[i-1][j-len[i]] + sum[j] - sum[j-len[i]]);34         }35         printf("%d\n", dp[m][n]);36     }37     return 0;38 }

Code Jun

 

HDU 1244 Max sum plus