int quickpow(int m,int n,int k){ int b = 1; while (n > 0) { if (n & 1) b = (b*m)%k; n = n >> 1 ; m = (m*m)%k; } return b;} 用2进制拆分理解这段代码
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int quickpow(int m,int n,int k){ int b = 1; while (n > 0) { if (n & 1) b = (b*m)%k; n = n >> 1 ; m = (m*m)%k; } return b;} 用2进制拆分理解这段代码
转载于:https://www.cnblogs.com/zy691357966/p/5480478.html
