hdu---(3555)Bomb(数位dp(入门))
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7921 Accepted Submission(s): 2778
counter-terrorists found a time bomb in the dust. But this time the
terrorists improve on the time bomb. The number sequence of the time
bomb counts from 1 to N. If the current number sequence includes the
sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
first line of input consists of an integer T (1 <= T <= 10000),
indicating the number of test cases. For each test case, there will be
an integer N (1 <= N <= 2^63-1) as the description.
The input terminates by end of file marker.
1
50
500
1
15
From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",
so the answer is 15.

//#define LOCAL
#include<cstdio>
#include<cstring>
#define LL __int64
using namespace std;
const int maxn=;
LL dp[maxn][]={};
int nn[maxn];
int main()
{ #ifdef LOCAL
freopen("test.in","r",stdin);
#endif
int cas,i;
LL n;
scanf("%d",&cas);
/*数位DP的惯有模式预处理*/
dp[][]=;
for(i=;i<=;i++)
{
dp[i][]=dp[i-][]*-dp[i-][];
dp[i][]=dp[i-][];
dp[i][]=dp[i-][]*+dp[i-][];
}
while(cas--)
{
scanf("%I64d",&n);
i=;
n+=;
memset(nn,,sizeof(nn));
while(n>)
{
nn[++i]=n%;
n/=;
}
LL ans=;
bool tag=;
int num=;
for( ; i>= ; i-- )
{
ans+=dp[i-][]*nn[i]; /*计算49开头的个数*/
if(tag){
ans+=dp[i-][]*nn[i]; /*当前面出现了49的时候,那么后面出现的任何数字也要进行统计*/
}
if(!tag&&nn[i]>)
{
ans+=dp[i-][]; /*如果没有出现49开头,只要首部大于5,那么必定保函有一个49*/
}
if(num==&&nn[i]==)
tag=;
num=nn[i];
}
printf("%I64d\n",ans);
}
return ;
}
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