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cf_467C_George and Job
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 choosek pairs of integers:
in such a way that the value of sum is maximal possible. Help George to cope with the task.
The first line contains three integers n,m and k(1?≤?(m?×?k)?≤?n?≤?5000). The second line containsn integers p1,?p2,?...,?pn(0?≤?pi?≤?109).
Print an integer in a single line — the maximum possible value of sum.
5 2 1 1 2 3 4 5
9
7 1 3 2 10 7 18 5 33 0
61
要求区间元素和,当然想求出数列的前缀和来了。
二维dp,dp[i][j]表示在取到第i个元素的时候取j个m长度的区间之和的最大值。
dp[i][j]可由dp[i-1][j]不添加区间得到的,或者由dp[i-m][j-1]添加一个区间得到。
遂有状态转移方程:dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+s[i]-s[i-m]),其中s是前缀和。
然后就是简单题了。
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; long long dp[5005][5005]; int main(){ int n,m,k; scanf ("%d%d%d",&n,&m,&k);// long long a[5005],s[5005]; memset(a,0,sizeof(a)); memset(s,0,sizeof(s)); memset(dp,0,sizeof(dp)); for (int i=1;i<=n;i++){ scanf ("%I64d",&a[i]); s[i]=s[i-1]+a[i]; } for (int i=m;i<=n;i++){ for (int j=1;j<=k;j++){ dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+s[i]-s[i-m]); } } printf ("%I64d\n",dp[n][k]); return 0; }
(ps:数据爆int) :(
cf_467C_George and Job