HDU 1061 rightmost digit problem solving method

Find the rightmost number of the value of a large number of n^n, that is, the lowest bit.

It will be able to solve a simple dichotomy.

Just pay attention to overflow, just put n% 10 after. I will not overflow, instead of using a long long.

#include <stdio.h>int rightmost (int n, int n) {if (n = = 0) return 1;int t = rightmost (N/2, n); t = T * t% 10;; if (n% 2) t *= n;return t% 10;} int main () {int T, n;scanf ("%d", &t), while (t--) {scanf ("%d", &n);p rintf ("%d\n", rightmost (n, n%10));} return 0;}

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HDU 1061 rightmost digit problem solving method