HDU 1087 Super Jumping! Jumping! Jumping!
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这个题一看就是找一个 递增子序列。。 然后脑残的以为找 最长上升子序列。 在模板里面加上个 sum 数组递推上 过了样例就交上去了
获得了 WA之后。 自己出了一组数据 3 2 1 8 9 这五个数 最长上升子序列 有三个 3 89 或者 2 8 9 或者 1 8 9 然后就没法处理。
后来一想 应该是 上升子序列中找和最大的。 就写出 n^2 的递推公式了。 s【i】 = max (s【i】,s【j】 + a【i】) 其中 s【i】 代表以第i个数为结尾的最大和。
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>#include <fstream>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <vector>#include <cmath>#include <iomanip>typedef long long LL;typedef unsigned long long LLU;const double PI=acos(-1.0);using namespace std;#define MAXN 1000+10#define INF 1 << 30int main (){ int n; while(scanf("%d",&n) && n){ LL a[MAXN]; LL s[MAXN] = {0}; for(int i = 1; i <= n; i++){ scanf("%I64d",&a[i]); s[i] = a[i]; } LL M = 0; for(int i = 1; i <= n; i++){ for(int j = 1; j <= i; j++){ if(a[i] > a[j]) s[i] = max(s[i], s[j]+a[i]); } } for(int i = 1; i <= n; i++) M = max(M, s[i]); printf("%I64d\n",M); } return 0;}
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- hdu/hdoj 1087 Super Jumping! Jumping! Jumping!
- HDU 1087 Super Jumping! Jumping! Jumping!
- hdu 1087 Super Jumping! Jumping! Jumping!
- hdu 1087 Super Jumping! Jumping! Jumping!
- HDU 1087 Super Jumping! Jumping! Jumping!
- hdu 1087 Super Jumping! Jumping! Jumping!
- HDU 1087 Super Jumping! Jumping! Jumping!
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- HDU 1087 Super Jumping! Jumping! Jumping!
- hdu 1087 Super Jumping! Jumping! Jumping!
- Hdu 1087 Super Jumping! Jumping! Jumping!
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- HDU 1087 Super Jumping! Jumping! Jumping!
- hdu 1087 Super Jumping! Jumping! Jumping!
- HDU 1087 Super Jumping! Jumping! Jumping!
- hdu 1087 Super Jumping! Jumping! Jumping!
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