uva 11029 - Leading and Trailing(快速幂)
2013-10-26 08:02
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题目链接:uva 11029 - Leading and Trailing
题目大意:给出一个n和k求n^k的前三位数和后三位数。
解题思路:后三为数可以用分治的方法(快速幂)去做,可是前三位数就比较麻烦了,看了别人的题解.
n^k = 10 ^ (k * log10(n)),所以可以将多余的位数移到小数点后面然后舍弃掉,只保留前三位,pow(10, 2 + fmod(k * log10(n), 1)).
题目大意:给出一个n和k求n^k的前三位数和后三位数。
解题思路:后三为数可以用分治的方法(快速幂)去做,可是前三位数就比较麻烦了,看了别人的题解.
n^k = 10 ^ (k * log10(n)),所以可以将多余的位数移到小数点后面然后舍弃掉,只保留前三位,pow(10, 2 + fmod(k * log10(n), 1)).
#include <stdio.h> #include <math.h> int n, k; long long pow_mod(long long a, long long c) { if (c == 1) return a % 1000; long long ans = pow_mod(a, c / 2); ans = (ans * ans) % 1000; return c % 2 ? (ans * a) % 1000 : ans; } void solve() { int l = (int) pow(10, 2 + fmod(k * log10(n), 1)); int r = pow_mod(n, k); printf("%d...%03d\n", l, r); } int main () { int cas; scanf("%d", &cas); while (cas--) { scanf("%d%d", &n, &k); solve(); } return 0; }
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