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动态规划第一题 值得纪念!~#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
int num[10000];
int dp[10000];
int main()
{
int n;
int i,j;
int len;
int max;
int finalmax;
int flag=1;
while(scanf("%d",&n))
{
if(n==-1)
break;
num[0]=n;
for(i=1;;i++)
{
scanf("%d",&n);
if(n==-1)
break;
else
num[i]=n;
}
len=i-1;
dp[len]=1;
for(i=len-1;i>=0;i--)
{
max=0;
for(j=i+1;j<=len;j++)
{
if(dp[j]>max&&num[j]<num[i])
max=dp[j];
}
dp[i]=max+1;
}
finalmax=0;
for(i=0;i<=len;i++)
{
if(dp[i]>finalmax)
finalmax=dp[i];
}
printf("Test #%d:\n maximum possible interceptions: %d\n\n",flag,finalmax);
flag++;
}
return 0;
}
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