wa哭了，，t哭了，，还是看了题解。。。
| 8170436 | 2014-10-11 06:41:51 | njczy2010 | C - George and Job | GNU C++ | Accepted | 109 ms | 196172 KB |
| 8170430 | 2014-10-11 06:39:47 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 1 | 61 ms | 196200 KB |
| 8170420 | 2014-10-11 06:37:14 | njczy2010 | C - George and Job | GNU C++ | Runtime error on test 23 | 140 ms | 196200 KB |
| 8170348 | 2014-10-11 06:16:59 | njczy2010 | C - George and Job | GNU C++ | Time limit exceeded on test 11 | 1000 ms | 196200 KB |
| 8170304 | 2014-10-11 06:05:15 | njczy2010 | C - George and Job | GNU C++ | Time limit exceeded on test 12 | 1000 ms | 196200 KB |
| 8170271 | 2014-10-11 05:53:29 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 3 | 77 ms | 196200 KB |
| 8170266 | 2014-10-11 05:52:37 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 3 | 62 ms | 196200 KB |
| 8170223 | 2014-10-11 05:39:00 | njczy2010 | C - George and Job | GNU C++ | Time limit exceeded on test 12 | 1000 ms | 196100 KB |
| 8170135 | 2014-10-11 05:07:06 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 9 | 93 ms | 196100 KB |
| 8170120 | 2014-10-11 05:01:28 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 5 | 78 ms | 196100 KB |
1 second
256 megabytes
standard input
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.
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).
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
转移方程为
dp[i][j]=max(dp[i+1][j],dp[i+m][j-1]+sum[i]); （转自：http://www.tuicool.com/articles/m6v6z23） C:显然是dp,设f[i][j]表示第i段的结尾为j时的最优值，显然f[i][j]=max{f[i-1][k]+sum[j]-sum[j-m]}(0<=k<=j-m) （转自：http://www.bkjia.com/ASPjc/881176.html）
不过这样是O(k*n^2),可能超时。
我们发现每一阶段的转移能用到的最优状态都是上一阶段的前缀最优值，于是dp时直接记录下来用来下面的转移，这样就不用枚举了，变为O(k*n)水过。
貌似卡内存，用了滚动数组。
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<string>
//#include<pair> #define N 5005
#define M 1000005
#define mod 1000000007
//#define p 10000007
#define mod2 100000000
#define ll long long
#define LL long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b) using namespace std; int n,m,k;
ll dp[N][N];
ll p[N];
ll ans;
ll sum[N]; void ini()
{
memset(dp,,sizeof(dp));
memset(sum,,sizeof(sum));
int i;
ll te=;
for(i=;i<=n;i++){
scanf("%I64d",&p[i]);
} for(i=n-m+;i<=n;i++){
te+=p[i];
}
sum[n-m+]=te;
dp[n-m+][]=te;
// ma=te;
for(i=n-m;i>=;i--){
sum[i]=sum[i+]+p[i]-p[i+m];
}
//for(i=1;i<=n;i++) printf(" i=%d sum=%I64d\n",i,sum[i]);
} void solve()
{
int i,j;
dp[n][]=sum[n];
for(i=n-;i>=;i--){
for(j=k;j>=;j--){
if(i+m<=n+)
dp[i][j]=max(dp[i+][j],dp[i+m][j-]+sum[i]);
else
dp[i][j]=dp[i+][j];
}
} } void out()
{
//for(int i=1;i<=n;i++){
// for(int j=0;j<=k;j++){
// printf(" i=%d j=%d dp=%I64d\n",i,j,DP(i,j));
// }
// }
printf("%I64d\n",dp[][k]);
} int main()
{
// freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
// scanf("%d",&T);
// for(int ccnt=1;ccnt<=T;ccnt++)
// while(T--)
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
//if(n==0 && k==0 ) break;
//printf("Case %d: ",ccnt);
ini();
solve();
out();
} return ;
}