http://acm.hdu.edu.cn/showproblem.php?pid=3555 (题目链接)
题意
求区间${[1,n]}$含有49的数的个数。
Solution
数位dp,先求出不含49的,再减一下就好了。
细节
LL
代码
// hdu3555
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<ctime>
#define LL long long
#define inf (1ll<<30)
#define MOD 1004535809
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
LL f[20][10],g[20],m,ans;
int t[20],n;
int main() {
int T;scanf("%d",&T);
while (T--) {
scanf("%lld",&m);ans=m;
for (n=0;m;m/=10) t[++n]=m%10;
memset(f,0,sizeof(f));memset(g,0,sizeof(g));
g[1]=1;
for (int i=0;i<10;i++) f[1][i]=1;
for (int i=2;i<=n;i++) {
for (int j=0;j<10;j++)
for (int k=0;k<10;k++)
if (j!=4 || k!=9) f[i][j]+=f[i-1][k];
for (int j=0;j<t[i-1];j++) if (t[i]!=4 || j!=9) g[i]+=f[i-1][j];
if (t[i]!=4 || t[i-1]!=9) g[i]+=g[i-1];
}
for (int i=1;i<n;i++)
for (int j=1;j<10;j++) ans-=f[i][j];
for (int i=1;i<t[n];i++) ans-=f[n][i];
ans-=g[n];
printf("%lld
",ans);
}
return 0;
}