uva 11361 - Investigating Div-Sum Property(數位dp)

標籤：

題意：題目大意：給出a，b，k，問說在[a,b]這個區間有多少n，滿足n整除k，以及n的各個為上的數字之和也整除k。

分析：dp[i][nmod][smod]長度為i,該數對k的餘數，各位和對k的餘數。

#include <map>#include <set>#include <list>#include <cmath>#include <queue>#include <stack>#include <cstdio>#include <vector>#include <string>#include <cctype>#include <complex>#include <cassert>#include <utility>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef pair<int,int> PII;typedef long long ll;#define lson l,m,rt<<1#define pi acos(-1.0)#define rson m+1,r,rt<<11#define All 1,N,1#define read freopen("in.txt", "r", stdin)const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;const int INF= 0x7ffffff;const int mod =  1000000007;int a,b,k;int dp[35][110][110],bit[35];int dfs(int i,int nmod,int smod,int e){    if(i==0)return (nmod==0)&&(smod==0);    if(!e&&dp[i][nmod][smod]!=-1)return dp[i][nmod][smod];    int l=e?bit[i]:9;    int num=0;    for(int v=0;v<=l;++v){        int tnmod=(nmod*10+v)%k;        int tsmod=(smod+v)%k;        num+=dfs(i-1,tnmod,tsmod,e&&(v==l));    }    return e?num:dp[i][nmod][smod]=num;}int solve(int x){    int len=0;    memset(dp,-1,sizeof(dp));    while(x){        bit[++len]=x%10;        x/=10;    }    return dfs(len,0,0,1);}int main(){    int t;    scanf("%d",&t);    while(t--){     scanf("%d%d%d",&a,&b,&k);        if(k>=100)           printf("0\n");        else{            printf("%d\n",solve(b)-solve(a-1));        }    }return 0;}

 

uva 11361 - Investigating Div-Sum Property(數位dp)