UVA 12105 - Bigger is Better(DP+高精度)
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题目链接:12105 - Bigger is Better
题意:一些火柴,问你能组成整除m最大的数字是多少。
思路:dp[i][j]表示用i根火柴,组成%m余数为j的最大数字,末尾多一个数字k的状态就是dp[i + num[k]][(j * 10 + k) % m],由于最多可能50位数,所以要用高精度。
注意一个优化点,由于高精度的计算上只需要乘10+k,常规的高精度乘法复杂度还是有点高会超时,所以用数组去模拟,每次*10 + k的时候就往后多一位即可。
代码:
#include <stdio.h>#include <string.h>const int N = 105;const int M = 3005;const int num[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};int n, m;char dp[N][M][55], ans[55], tmp[55];void init() {scanf("%d", &m);memset(dp, 0, sizeof(dp));memset(ans, 0, sizeof(ans));dp[0][0][0] = '0';}int numcmp(char *a, char *b) {int lena = strlen(a);int lenb = strlen(b);if (lena < lenb) return -1;if (lena > lenb) return 1;return strcmp(a, b);}void DP(char *a, char *b, int k) {memset(tmp, 0, sizeof(tmp));strcpy(tmp, b);int len = strlen(b);if (tmp[0] == '0')tmp[0] = '0' + k;elsetmp[len] = '0' + k;if (numcmp(a, tmp) < 0)strcpy(a, tmp);}int main() {int cas = 0;while (~scanf("%d", &n) && n) {init();for (int i = 0; i <= n; i++) {for (int j = 0; j < m; j++) {if (strlen(dp[i][j]) == 0) continue;for (int k = 0; k < 10; k++) {if (i + num[k] > n) continue;DP(dp[i + num[k]][(j * 10 + k) % m], dp[i][j], k);}}if (i != 0 && numcmp(ans, dp[i][0]) < 0)strcpy(ans, dp[i][0]);}printf("Case %d: ", ++cas);if (ans[0] == 0) printf("-1\n");else printf("%s\n", ans);}return 0;}
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