Uva 11361 Investigating Div-Sum Property(数位DP)

大致题意：

问在区间[a,b]内有多少个整数满足被K整除，且这个数的每一位上数字的和也被K整除

a ， b <= 2^31

思路：

显然在int范围内，每位数字之和不超过9*10 ，所以K大于100就直接返回0， 其余用数位dp算

数位dp的大致思路都一样，先dp出dp[len][x][y] 表示len位长的数有多少个数modK等于x，每一位和modK等于y

然后求DP(int num) 表示 不超过num 有多少个满足题意的数，输出DP(b) - DP(a-1) 就好

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
#define rep(i,n) for ( int i=0; i< int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
#define PB push_back
#define MP make_pair
typedef pair<int,int> pii;

template <class T>
inline bool RD(T &ret) {
        char c; int sgn;
        if (c = getchar(), c == EOF) return 0;
        while (c != '-' && (c<'0' || c>'9')) c = getchar();
        sgn = (c == '-') ? -1 : 1 , ret = (c == '-') ? 0 : (c - '0');
        while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
        ret *= sgn;
        return 1;
}
template <class T>
inline void PT(T x) {
        if (x < 0) putchar('-') ,x = -x;
        if (x > 9) PT(x / 10);
        putchar(x % 10 + '0');
}
const int N = 16;
int a, b, K;
ll dp[N][123][123]; // x - > *mod; y - > summod
ll p[N];
int bit[N];

ll DP(int x) {
	int t = x;
	int top = 0;
	ll ans = 0;
	int sum = 0;
	while( t ) bit[++top] = t % 10 , t /= 10 , sum += bit[top];
	ll curx = 0, cury = 0;
	for(int i = top; i >= 1; i -- ) {
		for(int dig = 0; dig < bit[i]; dig ++ ) {
			ll nx = ( K - ( curx + dig * p[i] ) % K ) % K;
			ll ny = ( K - ( cury + dig ) % K ) % K;
			ans += dp[i-1][nx][ny];
		}
		curx = (curx + bit[i] * p[i] ) % K;
		cury = (cury + bit[i] ) % K;		
	}
	return ans + ( sum % K == 0 && x % K == 0 );
}
int main() {
	p[1] = 1;
	for(int i = 2; i < N; i ++ ) p[i] = 10 * p[i-1];

	int T;
	RD(T);
	while(T --) {
		memset( dp, 0, sizeof(dp)) ;
		scanf("%d%d%d", &a, &b, &K);
		if( K >= 100 ) {
			puts("0"); 
			continue;
		}
		dp[0][0][0] = 1;
		rep(i, 15 ) {
			rep(j, K ) 
			rep( k, K ) {
				rep(dig, 10) {
					ll x = ( dig * p[i + 1] + j ) % K;
				   	ll y = ( dig + k ) % K ;
					dp[i+1][x][y] += dp[i][j][k] ;
				}	
			}
		}
		PT( DP(b) - DP(a - 1) );
		puts("");
	}
}