Bigger is Better

题意:
用n根火柴棒搭出数字,求这个数能被m整除的最大值
思路:递推的式子是newi=i+num[k][根数], newj=(j*10+k)%m[余数]
dp[newi][newj] = max(dp[i][j]+1, dp[newi][newj]);
那么ans是只针对余数为0的情况
出发点是dp[0][0],结尾点是dp[i][0],之间必定是会得出一条等差数列(d=1)铺成的路
所以就有dp[newi][newj]==d[i][j]+1, path[i][j] = k;
<pre name="code" class="cpp">#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 110;
const int maxm = 3010;
int dp[maxn][maxm], path[maxn][maxm];
int m, n;
int ans;
int num[10] = {6,2,5,5,4,5,6,3,7,6};
void deal() {for(int i=0; i<n; i++)for(int j=0; j<m; j++) {if(dp[i][j]>=0) //已经得出for(int k=9; k>=0; k--) {if(num[k]+i<=n){int newi = i+num[k];int newj = (j*10+k)%m;if(dp[i][j]+1>dp[newi][newj]) dp[newi][newj] = dp[i][j]+1;if(dp[newi][newj]>ans && newj==0) ans = dp[newi][newj];
}}}
}void find() {memset(path, -1, sizeof(path));for(int i=n; i>=0; i--)for(int j=0; j<m; j++)if(dp[i][j]>=0) {if(dp[i][j]==ans && j==0) {path[i][j] = 10;continue;}for(int k=9; k>=0; k--) if(i+num[k]<=n) {int newi = i+num[k], newj = (j*10+k)%m;if(dp[newi][newj]==(dp[i][j]+1) && path[newi][newj]>=0) {path[i][j] = k;break;}}}
}
void print() {int i, j, k, l;if(ans>0) {i = 0, j = 0;while(path[i][j] != 10) {k = i+num[path[i][j]];l = (j*10+path[i][j])%m;printf("%d", path[i][j]);i = k;j = l;}printf("\n");}else if(n>=num[0]) printf("0\n");else printf("-1\n");
}int main() {int kase = 1;while(scanf("%d", &n) != EOF && n) {scanf("%d", &m);printf("Case %d: ", kase++);memset(dp, -1, sizeof(dp));dp[0][0] = 0, ans = 0;deal();find();print();}return 0;
}

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