【数位dp】学习

根据 https://wenku.baidu.com/view/1e6ed0bbfd0a79563c1e72a8.html
发现数位dp不仅能做数位统计，也能做数位计算

//#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;typedef long long LL;#define mem(a, b) memset(a, b, sizeof(a))/*例一： 按位求和问题给定A，B（1<=A,B<=10^5）,求[A,B]内的所有数的k进制表示下各数位之和dp[i][j] 表示 0 ~ k^i-1 的位数和 （在j进制下）pw[i][j] 表示 j^i 的值对pw处理不理解 建议先做 51nod 1009题目没有提交，与网上代码对拍，无差异。*/#define Elem LLconst int Dight = 65;int bits[Dight];Elem dp[Dight][11];Elem pw[Dight][11];void init(){    for(int i = 2; i <= 10; i++)    {        pw[0][i] = 1;        for(int j = 1; j < Dight; j++)            pw[j][i] = pw[j-1][i]*(LL)i;    }}Elem dfs(int bit, int pos, int sta, bool limit){    if(pos==0)        return (LL)sta;    if(!limit && dp[pos][bit]!=-1)        return dp[pos][bit] + pw[pos][bit]*(LL)sta;    int up = limit ? bits[pos]:(bit-1);    Elem res = 0;    for(int i = 0; i <= up; i++)        res += dfs(bit,pos-1,sta+i,limit&&(i==up));    if(!limit)        dp[pos][bit] = res;    return res;}Elem solve(Elem n, int bit){    int pos = 0;    while(n){        bits[++pos] = n%bit;        n /= bit;    }    return dfs(bit,pos,0,1);}int main(){    init();    int T,k;    memset(dp,-1,sizeof(dp));//    freopen("G:\\duipai\\in.txt","r", stdin);//    freopen("G:\\duipai\\out1.txt","w", stdout);    scanf("%d",&T);    LL a,b;    while(T--){        scanf("%I64d%I64d%d",&a,&b,&k);        printf("%I64d\n",solve(b,k)-solve(a-1,k));    }    return 0;}