本文主要是介绍uva 748 Exponentiation（高精度实数乘法），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：748 - Exponentiation
题目大意：高精度求次幂。
解题思路：先将小树点剔除，记录小数点的位数。然后进行高精度乘法运算（注意输出时0的剔除）
#include <stdio.h>
#include <string.h>const int N = 1005;
int n, a[N], b[N], na, nb;int find(char *str) {int len = strlen(str);for (int i = len - 1; i >= 0; i--) if (str[i] == '.') return len - i - 1;return 0;
}void init(char *num) {na = nb = 0;memset(a, 0, sizeof(a));memset(b, 0, sizeof(b));int len = strlen(num);for (int i = len - 1; i >= 0; i--) if (num[i] != '.') a[na++] = num[i] - '0';b[0] = nb = 1;
}void put(int *num, int k, int cnt) {char str[N];int r = 0, l = 0;for (int i = cnt - 1; i >= 0; i--) {if (i == k - 1) str[r++] = '.';str[r++] = '0' + num[i];}while (str[l] == '0') l++;while (str[r - 1] == '0') r--;if (str[r - 1] == '.') r--;for (int i = l; i < r; i++) printf("%c", str[i]);printf("\n");
}void mul(int* x, int* y, int& nx, int& ny) {int sum, ans[N], tmp = 0;memset(ans, 0, sizeof(ans));for (int i = 0; i < nx; i++) {sum = 0;for (int j = 0; j < ny; j++) {sum = x[i] * y[j] + ans[i + j] + sum;ans[i + j] = sum % 10;sum /= 10;}int t = i + ny;while (sum) {ans[t++] = sum % 10;sum /= 10;}if (t > tmp) tmp = t;}nx = tmp;memcpy(x, ans, sizeof(ans));
}void solve(int k) {for (int i = 0; i < n; i++) mul(b, a, nb, na);put(b, k, nb);
}int main () {char num[N];while (scanf("%s%d", num, &n) == 2) {int k = find(num); k *= n;init(num);solve(k);}return 0;
}
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