3 0 100 1 10 5 100
Case #1: 1 Case #2: 2 Case #3: 13
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 10005
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int n,m,ans,cnt,tot,flag;
int a,b,fac[10];
int dp[10][10][5500],sum[10][10][5500];
void solve()
{
int i,j,t=0,u,v;
for(i=9;i>=1;i--)
{
t+=((a/fac[i-1])%10)*(1<<i-1);
}
a=t;
ans=0;
for(i=9;i>=1;i--)
{
u=(b/fac[i-1])%10;
for(j=0;j<u;j++)
{
ans+=sum[i][j][a];
}
a-=u*(1<<i-1);
if(a<0) return ;
}
ans++;
}
int main()
{
int i,j,k,p,q,t,test=0;
fac[0]=1;
for(i=1;i<10;i++)
{
fac[i]=fac[i-1]*10;
}
dp[0][0][0]=1;
for(i=1;i<=9;i++)
{
for(j=0;j<=9;j++)
{
for(k=0;k<=5000;k++)
{
t=0;
for(p=0;p<=9;p++)
{
if((k-(1<<i-1)*j)>=0) t+=dp[i-1][p][k-(1<<i-1)*j];
}
dp[i][j][k]=t;
if(k>0) sum[i][j][k]=sum[i][j][k-1]+t;
else sum[i][j][k]=t;
}
}
}
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&a,&b);
solve();
printf("Case #%d: %d\n",++test,ans);
}
return 0;
}
/*
3
0 100
1 10
5 100
Case #1: 1
Case #2: 2
Case #3: 13
*/
hdu 4734 F(x) (数位dp),布布扣,bubuko.com
原文:http://blog.csdn.net/tobewhatyouwanttobe/article/details/21295943