zoj 1539 Lot 简单DP 记忆化
2017-08-10 19:41
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Lot
Time Limit:
2 Seconds Memory Limit:
65536 KB
Out of N soldiers, standing in one line, it is required to choose several to send them scouting.
In order to do that, the following operation is performed several times: if the line consists of more than three soldiers, then all soldiers, standing on even positions, or all soldiers, standing on odd positions, are taken away. The above is done until three
or less soldiers are left in the line. They are sent scouting. Find, how many different groups of three scouts may be created this way.
Note: Groups with less than three number of soldiers are not taken into consideration.
0 < N <= 10 000 000
Input
The input file contains the number N.
Process to the end of file.
Output
The output file must contain the solution - the amount of variants.
Sample Input
10
4
Sample Output
2
0
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int f[10000001];
int _find(int v)
{
if(v<3) return f[v]=0;
if(v==3) return f[v]=1;
if(f[v]>0) return f[v];
if(v%2==0) return f[v]=_find(v/2)*2;
else return f[v]=_find(v/2)+_find(v/2+1);
}
int main()
{
int n,s;
while(cin>>n)//不知道为什么,这里换成输入优化就超超时了。。。。。。求大佬告知
{
s=_find(n);
if(s>0)
printf("%d\n",s);
else printf("0\n");
}
return 0;
}
Time Limit:
2 Seconds Memory Limit:
65536 KB
Out of N soldiers, standing in one line, it is required to choose several to send them scouting.
In order to do that, the following operation is performed several times: if the line consists of more than three soldiers, then all soldiers, standing on even positions, or all soldiers, standing on odd positions, are taken away. The above is done until three
or less soldiers are left in the line. They are sent scouting. Find, how many different groups of three scouts may be created this way.
Note: Groups with less than three number of soldiers are not taken into consideration.
0 < N <= 10 000 000
Input
The input file contains the number N.
Process to the end of file.
Output
The output file must contain the solution - the amount of variants.
Sample Input
10
4
Sample Output
2
0
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int f[10000001];
int _find(int v)
{
if(v<3) return f[v]=0;
if(v==3) return f[v]=1;
if(f[v]>0) return f[v];
if(v%2==0) return f[v]=_find(v/2)*2;
else return f[v]=_find(v/2)+_find(v/2+1);
}
int main()
{
int n,s;
while(cin>>n)//不知道为什么,这里换成输入优化就超超时了。。。。。。求大佬告知
{
s=_find(n);
if(s>0)
printf("%d\n",s);
else printf("0\n");
}
return 0;
}
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