链接：
https://vjudge.net/problem/HDU-3555
题意：
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?
思路:
考虑没有49的数，用a-掉就行。
Dp(i, j)
i位置为j时的值。
因为计算了0要多加1.
代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MOD = 1e9+7;
const int MAXN = 1e6+10;
LL Dp[25][10];
int dig[25];
LL a;
LL Dfs(int pos, int pre, int zer, int lim)
{
if (pos == -1)
return 1;
if (!lim && Dp[pos][pre] != -1)
return Dp[pos][pre];
int up = lim ? dig[pos] : 9;
LL cnt = 0;
for (int i = 0;i <= up;i++)
{
if (pre == 4 && i == 9)
continue;
cnt += Dfs(pos-1, i, zer && i == 0, lim && i == up);
}
if (!lim)
Dp[pos][pre] = cnt;
return cnt;
}
LL Solve(LL x)
{
int p = 0;
while(x)
{
dig[p++] = x%10;
x /= 10;
}
return Dfs(p-1, -1, 1, 1);
}
int main()
{
// freopen("test.in", "r", stdin);
int t;
scanf("%d", &t);
memset(Dp, -1, sizeof(Dp));
while(t--)
{
scanf("%lld", &a);
printf("%lld\n", a-Solve(a)+1);
}
return 0;
}