Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 16713 Accepted Submission(s): 6128
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
/*
数位DP.
f[i][j]表示以j开头的i位数的决策.
倒序存储
最后还要加一位统一开区间处理.
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#define MAXN 25
#define LL long long
using namespace std;
LL f[MAXN][MAXN],ans,n;
int t,l,a[MAXN];
void pre()
{
f[0][0]=1;
for(int i=1;i<=20;i++)//枚举位数.
for(int j=0;j<=9;j++)//枚举这一位数.
for(int k=0;k<=9;k++)//枚举上一位数.
{
if(j==4&&k==9) continue;
f[i][j]+=f[i-1][k];
}
return ;
}
void slove()
{
n++;ans=n;
l=0;memset(a,0,sizeof a);
while(n) a[++l]=n%10,n/=10;
a[l+1]=0;
for(int i=l;i;i--)
{
for(int j=0;j<a[i];j++)
{
if(a[i+1]==4&&j==9) continue;
ans-=f[i][j];
}
if(a[i]==9&&a[i+1]==4) break;
}
cout<<ans<<endl;
return ;
}
int main()
{
pre();scanf("%d",&t);
while(t--) cin>>n,slove();
return 0;
}