Hdu 4389 X mod f(x) 数位DP

X mod f(x)

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3177    Accepted Submission(s): 1234

Problem Description

Here is a function f(x):　　　int f ( int x ) {　　　    if ( x == 0 ) return 0;　　　    return f ( x / 10 ) + x % 10;　　　}

　　　Now, you want to know, in a given interval [A, B] (1 <= A <= B <= 10⁹), how many integer x that mod f(x) equal to 0.

 

Input

　　　The first line has an integer T (1 <= T <= 50), indicate the number of test cases.
　　　Each test case has two integers A, B.

 

Output

　　　For each test case, output only one line containing the case number and an integer indicated the number of x.

 

Sample Input

21 1011 20

 

Sample Output

Case 1: 10Case 2: 3

 

Author

WHU

 

Source

2012 Multi-University Training Contest 9

这题和BZOJ 1799 : http://blog.csdn.net/sinat_35406909/article/details/75808465

题意一样。

只不过，这题由于内存限制比较大，可以直接多一维存储搜索过的状态。

一样的穷举数位和进行DP。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <bitset>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;const int maxn=82,inf=0x3f3f3f3f;const ll llinf=0x3f3f3f3f3f3f3f3f; int dp[11][maxn][maxn][maxn];int num[20];int dfs(int len,int sum,int mod,int nu,bool HaveLimit) {    if (sum>len*9) return 0;    if (len==0)         return sum==0&&nu==0;    if (dp[len][sum][mod][nu]!=-1&&!HaveLimit)         return dp[len][sum][mod][nu];            int p=HaveLimit?num[len]:9;    int ans=0;    for (int i=0;i<=p;i++) {        ans+=dfs(len-1,sum-i,mod,(nu*10+i)%mod,HaveLimit&&i==num[len]);    }    if (!HaveLimit)         dp[len][sum][mod][nu]=ans;    return ans;}int solve(int n) {    int len=0,i;    int k=n;    while (k) {        num[++len]=k%10;        k/=10;    }    int ans=0;    for (i=1;i<=min(9*len,81);i++) {        ans+=dfs(len,i,i,0,1);    }    return ans;}int main() {    int a,b;    int cas,cnt=0;    memset(dp,-1,sizeof(dp));    scanf("%d",&cas);    while (scanf("%d%d",&a,&b)!=EOF) {        cnt++;        int ans=solve(b)-solve(a-1);        printf("Case %d: %d\n",cnt,ans);    }    return 0;}