UVa 11361 Investigating Div-Sum Property / 数位DP

先上代码 以后再说

#include <cstdio>
#include <cstring>
const int maxn = 110;
int dp[maxn][maxn][maxn];
int ok(int x, int k)
{
	if(x < 10)
		return x / k;
	int a = x;
	int b = 1;
	int l = 0;
	while(a)
	{
		l++;
		a /= 10;
		b *= 10;
	}
	b /= 10;
	//printf("%d %d\n", a, b);
	int ret = 0;
	int m1 = 0, m2 = 0;
	int d = 1;
	while(l--)
	{
		int c = x / b;
		//printf("%d\n", c);
		if(l == 0)
		{
			for(int j = 0; j <= c; j++)
				if(((k-j-m1)%k+k)%k == 0 && ((k-j-m2)%k+k)%k == 0)
					ret++;
			break;
		}
		for(int j = 0; j < c; j++)
			ret += dp[l-1][((k-j-m1)%k+k)%k][((k-j*b-m2)%k+k)%k];
		//	printf("%d\n", c);
		m1 += c;
		m2 += c*b;
		x -= c*b;
		b /= 10;
	}
	return ret-1;
}
int main()
{
	int T;
	int a, b, k;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d %d %d", &a, &b, &k);
		if(k > 82)
		{
			printf("0\n");
			continue;
		}
		int c = a, d = 1;
		memset(dp, 0, sizeof(dp));
		for(int i = 0; i < 10; i++)
			dp[0][i%k][i%k]++;
		for(int i = 1; i <= 10; i++)
		{
			d *= 10;
			if(d * 10 > b)
			{
				//printf("%d\n", d);
				break;
			}
			for(int m1 = 0; m1 < k; m1++)
			{
				for(int m2 = 0; m2 < k; m2++)
				{
					for(int j = 0; j < 10; j++)
					{
						dp[i][((m1+j)%k+k)%k][((m2+j*d)%k+k)%k] += dp[i-1][m1][m2];
					}
				}
			}
		}	
		printf("%d\n", ok(b,k)-ok(a-1,k));
	}
	return 0;
}

UVa 11361 Investigating Div-Sum Property / 数位DP

原文：http://blog.csdn.net/u011686226/article/details/18883851