lightOJ 1068 - Investigation (数位dp)

1068 - Investigation
PDF (English) Statistics Forum
Time Limit: 2 second(s) Memory Limit: 32 MB

An integer is divisible by 3 if the sum of its digits is also divisible by 3. For example, 3702 is divisible by 3 and 12 (3+7+0+2) is also divisible by 3. This property also holds for the integer 9.

In this problem, we will investigate this property for other integers.
Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains three positive integers A, B and K (1 ≤ A ≤ B < 231 and 0 < K < 10000).
Output

For each case, output the case number and the number of integers in the range [A, B] which are divisible by K and the sum of its digits is also divisible by K.
Sample Input

Output for Sample Input

3

1 20 1

1 20 2

1 1000 4

Case 1: 20

Case 2: 5

Case 3: 64

这种题目考验智商！！
n大于82的时候是没用的，因为数位之和小于82.
还有，k在变化
所以每次都要memset

#include<iostream>using namespace std;#include<cstring>#include<cstdio>#include<algorithm>int digit[20];int k;int dp[10][83][83];int dfs(int pos,int lim,int dig,int sum) {    if(pos<0)return dig==0&&sum==0;    if(!lim&&(~dp[pos][dig][sum]))return dp[pos][dig][sum];    int ans=0;    int len=lim?digit[pos]:9;    for(int i=0; i<=len; ++i) {        ans+=dfs(pos-1,lim&&(i==len),((dig*10+i)%k)%83,(sum+i)%k);    }    if(!lim)dp[pos][dig][sum]=ans;    return ans;}int work(int val) {    int len=0;    while(val) {        digit[len++]=val%10;        val/=10;    }    return dfs(len-1,1,0,0);}int main() {    int n,m,T;    int coun=0;    scanf("%d",&T);    while(T--&&scanf("%d%d%d",&n,&m,&k)) {        memset(dp,-1,sizeof(dp));        printf("Case %d: %d\n",++coun,(work(m)-work(n-1)));    }    return 0;}