Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 18060 Accepted Submission(s): 6639
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?
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.
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
Hint
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.Author
fatboy_cw@WHU
Source
推荐一份题解 写的很好 点我
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<cstring>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
#include<string.h>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=0.0000000001;
const int N=100;
ll dp[N][N];
int a[N];
void init(){
memset(dp,0,sizeof(dp));
dp[0][0]=1;dp[0][1]=0;dp[0][2]=0;
for(int 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;
}
}
int main()
{
int T;
init();
scanf("%d",&T);
while(T--){
memset(a,0,sizeof(a));
ll n;
scanf("%I64d",&n);
n++;
int t=0;
while(n){
a[++t]=n%10;
n/=10;
}
a[t+1]=0;
int flag=0;
ll sum=0;
for(int i=t;i>=1;i--){
sum=sum+dp[i-1][2]*a[i];
if(flag==0&&a[i]>4) sum=sum+dp[i-1][1];
if(flag) sum=sum+dp[i-1][0]*a[i];
if(a[i+1]==4&&a[i]==9) flag=1;
}
cout<<sum<<endl;
}
return 0;
}