[luoguP1773] 符文之语_NOI导刊2010提高（02）（DP）

Posted 2020-10-02 蒟蒻zht的博客

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f[i][j]表示前i个数余数为j的最优解

sum[i][j]表示字符串i~j所构成的数

 

#include <cstdio>
#include <cstring>
#define N 1001
#define min(x, y) ((x) < (y) ? (x) : (y))

int n, p;
char s[N];
int f[N][51], sum[N][N];

int main()
{
	int i, j, k;
	scanf("%s", s + 1);
	n = strlen(s + 1);
	scanf("%d", &p);
	for(i = 1; i <= n; i++)
		for(j = i; j <= n; j++)
			sum[i][j] = (sum[i][j - 1] * 10 + s[j] - ‘0‘) % p;
	memset(f, 127 / 3, sizeof(f));
	for(i = 1; i <= n; i++) f[i][sum[1][i]] = 0;
	for(i = 1; i <= n; i++)
		for(j = 0; j < p; j++)
			for(k = 1; k < i; k++)
				f[i][j * sum[k + 1][i] % p] = min(f[i][j * sum[k + 1][i] % p], f[k][j] + 1);
	for(i = 0; i < p; i++)
		if(f[n][i] < 707406378)
		{
			printf("%d %d ", i, f[n][i]);
			break;
		}
	for(i = p - 1; i >= 0; i--)
		if(f[n][i] < 707406378)
		{
			printf("%d %d ", i, f[n][i]);
			break;
		}
	return 0;
}