HDU 4389 X mod f(x)

题解：f(x)表示x的各个数位的数字之和，求满足x在[a, b]之间且x = 0(mod f(x))的所有x的个数。

解法：数位DP，这道题思维难度一般，但是充分体现了数位DP锻炼代码的特性。

　　　设d[i][j][k][l]表示长度为i，各位数字之和为j，除以k余l的数的个数。具体状态转移方程见代码。

tag:数位DP

/*
* Author:  Plumrain
* Created Time:  2013-12-16 12:19
* File Name: DP-HDU-4389.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

#define CLR(x) memset(x, 0, sizeof(x))
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<#a<<endl
typedef long long int64;
int dit[20];
int64 ten[15];
int d[11][85][85][85];

void init()
{
ten[0] = 1;
for (int i = 1; i < 15; ++ i)
ten[i] = ten[i-1] * 10;

CLR (d);
for (int i = 0; i < 10; ++ i)
for (int j = 1; j < 85; ++ j)
++ d[1][i][j][i%j];

for (int i = 2; i < 10; ++ i)
for (int j = 0; j < 85; ++ j)
for (int k = 1; k < 85; ++ k)
for (int l = 0; l < k; ++ l)
for (int t = 0; t < 10; ++ t) if (j+t < 85)
d[i][j+t][k][(l+t*ten[i-1])%k] += d[i-1][j][k][l];
}

int gao(int x)
{
if (x < 10) return x-1;
int64 ret = 0, len = 0;
while (x){
dit[len++] = x % 10;
x /= 10;
}
dit[len] = 0;

int64 flag = 0, dt_sum = 0;
for (int i = len-1; i > 0; -- i){
for (int t = 0; t < dit[i]; ++ t)
for (int j = 0; j < 82; ++ j)
for (int l = 0; l < 82; ++ l){
int k = j + t + dt_sum;
if (!k || (flag + t*ten[i] + l) % k != 0) continue;
ret += d[i][j][k][l];
}

flag += dit[i] * ten[i];
dt_sum += dit[i];
}

for (int i = 0; i < dit[0]; ++ i)
if ((flag+i) % (dt_sum+i) == 0) ++ ret;
return (int)ret;
}

int main()
{
//freopen("a.in","r",stdin);
//freopen("my.out","w",stdout);
init();
int T, test = 0;
scanf ("%d", &T);
while (T--){
int a, b;
scanf ("%d%d", &a, &b);
printf ("Case %d: %d\n", ++test, gao(b+1) - gao(a));
}
return 0;
}

View Code