UV 12105-Bigger is Better (dp)

Source: Internet
Author: User

Link to the question: Ultraviolet A 12105-Bigger is Better

There are n matches, and a number can be divided by m, and the maximum.

Solution: dp [I] [j] indicates that I match is used to form a number that modulo m and j, but the State retains a string.

     using namespace std;const int N = 105;const int M = 3005;const int need[N] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};int n, m;char dp[N][M][55];bool cmp(char* a, char* b) {int la = strlen(a);int lb = strlen(b);if (la != lb)return la < lb;return strcmp(a, b) < 0;}void solve (char* a, char* b, int k) {char tmp[55];memset(tmp, 0, sizeof(tmp));strcpy(tmp, b);int len = strlen(tmp);if (tmp[0] == '0')tmp[0] = '0' + k;elsetmp[len] = '0' + k;if (cmp(a, tmp))strcpy(a, tmp);}int main () {int cas = 1;while (scanf("%d", &n) == 1 && n) {scanf("%d", &m);char ans[55];memset(ans, 0, sizeof(ans));memset(dp, 0, sizeof(dp));dp[0][0][0] = '0';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 + need[k] > n) continue;int t = (j*10 + k)%m;solve(dp[i+need[k]][t], dp[i][j], k);}}if (i && cmp(ans, dp[i][0]))memcpy(ans, dp[i][0], sizeof(dp[i][0]));}printf("Case %d: ", cas++);if (ans[0] == '\0')printf("-1\n");elseprintf("%s\n", ans);}return 0;}

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