BZOJ 1799 [Ahoi2009]self 同类分布 数位DP

1799: [Ahoi2009]self 同类分布

Time Limit: 50 Sec  Memory Limit: 64 MB
Submit: 1211  Solved: 517
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Description

给出a,b，求出[a,b]中各位数字之和能整除原数的数的个数。

Input

Output

Sample Input

10 19

Sample Output

3

HINT

【约束条件】1 ≤ a ≤ b ≤ 10^18

Source

Day1

穷举数位和

DP记录三个量：长度，数位和，当前数字除以数位和的余数。

至于剩下的，套一套模板。

不过这个DP的状态挺难想的。

#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 inf=0x3f3f3f3f;const ll llinf=0x3f3f3f3f3f3f3f3f; ll dp[20][165][165];int num[20];ll 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][nu]!=-1&&!HaveLimit)     return dp[len][sum][nu];    int p=HaveLimit?num[len]:9;ll 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][nu]=ans;return ans;}ll solve(ll n) {int len=0,i;ll k=n;while (k) {num[++len]=k%10;k/=10;}ll ans=0;for (i=1;i<=9*len;i++) {memset(dp,-1,sizeof(dp));ans+=dfs(len,i,i,0,1);}return ans;}int main() {ll a,b;while (scanf("%lld%lld",&a,&b)!=EOF) {ll ans=solve(b)-solve(a-1);printf("%lld\n",ans);}return 0;}