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uva 12105 - Bigger is Better(dp)

2014-05-05 22:33 453 查看
题目链接:uva 12105 - Bigger is Better

题目大意:有n根火柴,要组成一个数字能够整除m,并且最大。

解题思路:dp[i][j]表示用了i个火柴,组成的数字模掉m余j的情况,只不过状态保留的是字符串。

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>

using namespace std;
const int N = 105;
const int M = 3005;
const int need
= {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};

int n, m;
char dp
[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;
else
tmp[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");
else
printf("%s\n", ans);
}
return 0;
}
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