hdu 3555 Bomb（数位dp）

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 4054 Accepted Submission(s): 1401



Problem Description

The 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?

Input

The 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.

Output

For each test case, output an integer indicating the final points of the power.

Sample Input

3
1
50
500

Sample Output

0
1
15

HintFrom 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.

Author

fatboy_cw@WHU

Source

2010 ACM-ICPC Multi-University
Training Contest（12）——Host by WHU

Recommend

zhouzeyong

题意：找1~N的包含49的数的个数

#include<stdio.h>
#include<string.h>
long long dp[28][3],n,res;
int c[28];
int main()
{
int i,t,flag,len;

memset(dp,0,sizeof(dp));
for(dp[0][0]=i=1;i<=20;i++)
{
dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
dp[i][1]=dp[i-1][0];
dp[i][2]=dp[i-1][1]+dp[i-1][2]*10;
}
scanf("%d",&t);
while(t--)
{
scanf("%I64d",&n);
memset(c,0,sizeof(c));
res=flag=len=0;
n++;
while(n)
{
c[++len]=n%10;
n/=10;
}
for(i=len;i>=1;i--)
{
res+=dp[i-1][2]*c[i];
if(flag) res+=dp[i-1][0]*c[i];
if(!flag&&c[i]>4) res+=dp[i-1][1];
if(c[i+1]==4&&c[i]==9) flag=1;
}
printf("%I64d\n",res);
}

return 0;
}