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