HDU-3555 Bomb

Bomb

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
[/align]
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?

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, output an integer indicating the final points of the power.
 

[align=left]Sample Input[/align]

3
1
50
500

 

[align=left]Sample Output[/align]

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.

 ————————————————————集训20.1的分割线————————————————————

思路：水题一枚，数位DP，记录前一位是不是4，记录是否出现了"49"即可。注意使用%I64d

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <iostream>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
/****************************************/
int num[20];
LL dp[20][2][2];

LL dfs(int len, bool pre, bool bomb, bool bnd)
{
if(len == 0) {
if(bomb) return 1;
return 0;
}
if(!bnd && dp[len][pre][bomb] != -1)
return dp[len][pre][bomb];
int lim = (bnd ? num[len] : 9);
LL ret = 0;
for(int k = lim; k >= 0; k--) {
ret += dfs(len-1, k==4, bomb || (pre&&k==9), bnd && k==lim);
}
if(!bnd) dp[len][pre][bomb] = ret;
return ret;
}

LL Solve(LL x)
{
int cnt = 0;
while(x) {
num[++cnt] = x % 10;
x /= 10;
}
return dfs(cnt, false, false, true);
}

int main()
{
#ifdef J_Sure
// freopen("000.in", "r", stdin);
// freopen(".out", "w", stdout);
#endif
LL n;
int T;
memset(dp, -1, sizeof(dp));
scanf("%d", &T);
while(T--) {
scanf("%I64d", &n);
printf("%I64d\n", Solve(n));
}
return 0;
}