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?
InputThe 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.
OutputFor each test case, output an integer indicating the final points of the power.Sample Input
3 1 50 500Sample 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.
题意：
问小于等于n的数字中包含着49的数字有多少个。
做法：
数位dp板子题目，统计有多少个不包含49的，减去就好
代码：
#include<iostream>
using namespace std;
#include<cstdio>
#include<cstring>
#include<cstdlib>
typedef long long ll;
const int state = 2;
int a[25];
ll dp[25][state];
ll dfs(int pos,int pre,bool limit){
if(pos == -1)
return 1;//深度
int up;//枚举上界
if(!limit&&dp[pos][pre]!=-1)//记忆化
return dp[pos][pre];
up = limit?a[pos]:9;
ll ans = 0;
for(int i=0;i<=up;i++){
if(pre==1&&i==9)
continue;
ans+=dfs(pos-1,i==4,limit&&i==a[pos]);
}
if(!limit)//有选择地记忆化
return dp[pos][pre]=ans;
else
return ans;
}
ll solve(ll x){
int pos = 0;
memset(a,0,sizeof(a));
while(x){
a[pos++]=x%10;
x/=10;
}
return dfs(pos-1,0,true);
}
int main(){
ll p,q;
memset(dp,-1,sizeof(dp));
int t;
scanf("%d",&t);
while(t--){
ll s;
scanf("%lld",&s);
p=solve(s);
q = s - p + 1;
printf("%lld\n",q);
}
return 0;
}
这篇博客不错，学习数位DP的童鞋可以看一下：
