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51nod 1052 最大M子段和

2017-11-17 22:24 417 查看
dp[i]表示把前i个数分成x份的最大和

pre[i]表示把前i个数分成x-1份的最大和

进行m次dp

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ansn() printf("%d\n",ans)
#define lansn() printf("%lld\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000000+7;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
//const ll infl = 10000000000000000;
const int maxn=  5e3+10;
const int maxm = 2e4+10;
//Pretests passed
int in(ll &ret)
{
char c;
int sgn ;
if(c=getchar(),c==EOF)return -1;
while(c!='-'&&(c<'0'||c>'9'))c=getchar();
sgn = (c=='-')?-1:1;
ret = (c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
ret *=sgn;
return 1;
}
ll pre[maxn];
ll dp[maxn];
int num[maxn];
int main()
{
#ifdef LOCAL
freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

int n,m;
sdd(n,m);
for(int i=1;i<=n;++i)sd(num[i]),dp[i] = -inf;
for(int j = 1 ;j<=m;++j)
{
ll ans = -inf;
for(int i=j;i<=n;++i)
{
dp[i] = max(dp[i-1],pre[i-1]) + num[i];
pre[i-1] = ans;
ans = max(ans,dp[i]);
}
pre
= ans;
}
ll ans = pre
;
lansn();

return 0;
}
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