CodeForces - 467C George and Job

CodeForces - 467C George and Job Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description 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 Input Input 5 2 1 1 2 3 4 5 Output 9 Input 7 1 3 2 10 7 18 5 33 0 Output 61   #include<stdio.h> #include<string.h> #include<algorithm> #define N 100010 #define ll long long using namespace std; ll a[N]; ll dp[2][N]; ll sum[N]; int main() { int n,m,k; while(scanf("%d%d%d",&n,&m,&k)!=EOF) { ll ans=0; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); sum[i]=0; ans+=a[i]; if(i<m) continue; for(int j=i;j>i-m;j--) sum[i]+=a[j]; } if(n==k||n==m) { printf("%lld\n",ans); continue; } memset(dp,0,sizeof(dp)); ans=0; for(int l=1;l<=k;l++) { for(int i=l*m;i<=n;i++) { dp[l&1][i]=max(dp[l&1][i],dp[l&1][i-1]); dp[l&1][i]=max(dp[l&1][i],dp[(l-1)&1][i-m]+sum[i]); ans=max(ans,dp[l&1][i]); } } printf("%lld\n",ans); } return 0; }