lifehappy

D. Steps to One设f[i]f[i]f[i]为gcd⁡\gcdgcd为iii，还需要多少个数，那么有f[i]=1+∑j=1mf[gcd⁡(i,j)]mf[i] = 1 + \frac{\sum\limits_{j = 1} ^{m} f[\gcd(i, j)]}{m}f[i]=1+mj=1∑m​f[gcd(i,j)]​，f[1]=0f[1] = 0f[1]=0，考虑化简∑j=1mf[gcd⁡(i,j)]\sum\limits_{j = 1} ^{m} f[\gcd(i, j)]j=1∑m​f

概率期望LOOPSdp[i][j]dp[i][j]dp[i][j]表示从i,ji, ji,j到r,cr, cr,c的期望，有dp[i][j]=p0×dp[i][j]+p1×dp[i][j+1]+p2×dp[i+1][j]+2dp[i][j] = p_0 \times dp[i][j] + p_1 \times dp[i][j + 1] + p_2 \times dp[i + 1][j] + 2dp[i][j]=p0​×dp[i][j]+p1​×dp[i][j+1]+p2​×dp[i+1][j]+2有d