D - 稳住GCD DP - 相关文章

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UESTC 923 稳住GCD DP + GCD

定义:dp[i][j] 表示在前i个数中,使整个gcd值为j时最少取的数个数. 则有方程: gg = gcd(a[i],j) gg == j : 添加这个数gcd不变,不添加, dp[i][j] = dp[i-1][j] gg != j: t添加,更新答案, dp[i][gg] = dp[i-1][j] + 1 最后答案为dp[n][g] (g为原始的所有数的gcd) 时间复杂度: O(n*max(a[i])) 代码: #include <iostream>…

D - 稳住GCD DP

http://acm.uestc.edu.cn/#/problem/show/923 给定一堆数字,求其所有数字的gcd. 现在要删除最多的数字,使得剩下的数字的gcd和原来的一样. 设dp[i][val]表示在前i个数中,得到val这个数字所需的最小数字,怎么得到val这个数字?就是gcd得到. dp[i][gcd] 然后转移就是 dp[i][a[i]] = 1是必然的,自己一个枚举新数字得到新的gcd val dp[i][val] = min(dp[i][val], dp[i - 1][…

[原]携程预选赛A题-聪明的猴子-GCD+DP

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hdu CA Loves GCD(dp)

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