HDU-6156 Palindrome Function(数位DP)

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题意：

已知函数f(x, k)，如果10进制数x在k进制下是个回文数，那么f(x, k)值为k，否则为1。现给出L, R, l, r, 求出∑∑f(i, j) 其中(L<=i<=R)(l<=j<=r)

思路：

枚举每个进制，然后进行数位dp就好了，原本想少写点开个二维，但是没法记忆化了，就跟暴搜一样了= =，于是开了个思维。

dp[pos][i][j][k]表示第pos位i进制下回文串长度为j且字符串其上一位为k时候的记忆化值。

代码：

#include <bits/stdc++.h>#define LL long longusing namespace std;int L, R, l, r;int wei[65], now[65];LL dp[65][40][40][40];LL dfs(int pos, int limit, int lead, int bas, int len, int ft){if(pos == -1) return 1;if(!limit && !lead && dp[pos][bas][len][ft] != -1) return dp[pos][bas][len][ft];int up = limit? wei[pos]: bas-1;LL tmp = 0;for(int i = 0; i <= up; ++i){if(lead && i == 0) tmp += dfs(pos-1, limit&&i==wei[pos], 1, bas, 39, 39);else{if(lead){now[1] = i;tmp += dfs(pos-1, limit&&i==wei[pos], 0, bas, pos+1, i);}else{if(len-pos <= len/2){now[len-pos] = i;tmp += dfs(pos-1, limit&&i==wei[pos], 0, bas, len, ft);}else if(len&1 && len-pos-1 == len/2){tmp += dfs(pos-1, limit&&i==wei[pos], 0, bas, len, ft);}else if(len-pos > len/2){if(now[pos+1] != i) continue;tmp += dfs(pos-1, limit&&i==wei[pos], 0, bas, len, ft);}}}}if(!limit && !lead) dp[pos][bas][len][ft] = tmp;return tmp;}LL work(LL x, int bas){int pos = 0;while(x){wei[pos++] = x%bas;x /= bas;}return dfs(pos-1, 1, 1, bas, 39, 39);}int main(){memset(dp, -1, sizeof dp);int t; scanf("%d", &t);for(int _ = 1; _ <= t; ++_){LL tmp, ans = 0;scanf("%d %d %d %d", &L, &R, &l, &r);for(int i = l; i <= r; ++i){tmp = work(R, i)-work(L-1, i);ans += tmp*i + (R-L+1-tmp);}printf("Case #%d: %lld\n", _, ans);}return 0;}

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