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Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
//#include <bits/stdc++.h> #include <vector> #include <iostream> #include <queue> #include <cmath> #include <map> #include <cstring> #include <algorithm> #include <cstdio> using namespace std; #define Riep(n) for(int i=1;i<=n;i++) #define Riop(n) for(int i=0;i<n;i++) #define Rjep(n) for(int j=1;j<=n;j++) #define Rjop(n) for(int j=0;j<n;j++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const LL inf=1e14; const int N=1e4+15; const int maxn=18; int p[10*N],a[10*N]; LL dp[10*N]; int main() { int n; while(cin>>n) { LL sum=0,ans=0; Riep(n)read(a[i]),p[i]=(1890*a[i]+143)%10007-a[i],sum=sum+a[i]; dp[0]=0; for(int i=1;i<=n;i++) { if(dp[i-1]+p[i]<0)dp[i]=0; else dp[i]=dp[i-1]+p[i]; ans=max(ans,dp[i]); } cout<<sum+ans<<"\n"; } return 0; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5593733.html