51nod 1052 最大M子段和

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 passedint 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[n] = ans;    }    ll ans = pre[n];    lansn();    return 0;}