UVA11361 Investigating Div-Sum Property

Investigating Div-Sum Property

 题目大意：给定一个k,求a <= x <= b的，且x%k为0，x十进制下%k为0，有多少个x

 

题解：

数位dp套路题

ddp[i][j][a]表示i位数，%k为j，各位数加和%k为a的数有多少

初始化出i=1的状态，枚举往后面一位加的数b  dp[i+1][(j * 10 + b)%k][(j+b)%k] += dp[i][j][a]

 

直接求a与b之间的不好求，转为求1....b减去1....a-1

从高位到低位枚举，比如13245：
枚举0****，10***，11***，12***，130**，131**，1320*，1321*，1322*，1323*，分别用对应状态加入答案

然后枚举13240,13241,13242,13243,13244,13245，加入答案

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(long long &a, long long &b)
{
    long long tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const long long INF = 0x3f3f3f3f;
const long long MAXK = 100 + 10;

long long t,a,b,k,dp[40][MAXK][MAXK],powp[40],n,tmp;
//dp[i][j][a]表示i位数，这个数%k=j,各位数之和%k=a


//求1...m有多少满足条件的数 
long long solve(long long m)
{
    long long now = 0, sum = 0, ans = 0, ma = 0;
    while(powp[ma] <= m) ++ ma;
    for(register long long i = ma;i >= 2;-- i)
    {
        while(now + powp[i - 1] <= m) 
        {
            ans += dp[i - 1][(k - (now % k))%k][(k - (sum % k))%k];
            now += powp[i - 1];
            ++ sum;
        }
    }
    for(;now <= m;++ now, ++ sum)
        if(now % k == 0 && sum % k == 0) ++ ans;
    return ans;
}

int main()
{
    powp[0] = 1;
    for(register long long i = 1;i <= 35;++ i)powp[i] = powp[i - 1] * 10;
    read(t);
    for(;t;--t)
    {
        read(a), read(b), read(k);
        if(k > 100)
        {
            printf("0\n");
            continue;
        }
        memset(dp, 0, sizeof(dp));
        n = 0;
        while(powp[n] <= b) ++ n;
        for(register long long i = 0;i <= 9;++ i) ++ dp[1][i % k][i % k];
        for(register long long i = 1;i < n;++ i)
            for(register long long j = 0;j < k;++ j)
                for(register long long a = 0;a < k;++ a)
                    for(register long long b = 0;b <= 9;++ b)
                        dp[i + 1][(j * 10 + b) % k][(a + b) % k] += dp[i][j][a];
        printf("%lld\n", solve(b) - solve(a - 1));
    }
    return 0;
}