POJ 1001.Exponentiation

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题目:http://icpc.njust.edu.cn/Problem/Pku/1001/

AC代码(C++):

#include <iostream>#include <stdio.h>#include <algorithm>#include <iomanip>#include <string.h>using namespace std;int main(void) {char sample[7];int sn;int cn = 0;char rs[50][150];int ptn[50];for (int i = 0; i < 50; i++)ptn[i] = -1;while (cin >> sample >> sn) {reverse(sample, sample + 6);int pt = 0;for (; sample[pt] != '.'; pt++)if (pt == 5 && sample[pt] != '.') {pt = 0;break;}ptn[cn] = pt*sn - 1;if (pt != 0)for (int i = pt; i < 6; i++)sample[i] = sample[i + 1];for (int i = 0; i < 150; i++)rs[cn][i] = '\0';if (sn == 0) {cout << 1 << endl;continue;}if (sn == 1) {for (int j = 0; sample[j] != '\0'; j++) {rs[cn][j] = sample[j];if (sample[j + 1] == '\0')rs[cn][j + 1] = '\0';}}for (int i = 1; i < sn; i++) {char tr[6][150];for (int j = 0; j < 150; j++)for (int k = 0; k < 6; k++)tr[k][j] = '\0';if (i == 1)for (int j = 0; sample[j] != '\0'; j++) {rs[cn][j] = sample[j];if (sample[j + 1] == '\0')rs[cn][j + 1] = '\0';}for (int j = 0; sample[j] != '\0'; j++) {int t = 0;for (int k = 0; rs[cn][k] != '\0'; k++) {tr[j][k] = ((sample[j] - '0')*(rs[cn][k] - '0') + t) % 10 + '0';t = ((sample[j] - '0')*(rs[cn][k] - '0') + t) / 10;if (rs[cn][k + 1] == '\0'&&t != 0)tr[j][k + 1] = t + '0';}}int tt = 0;for (int j = 0; j < 150; j++) {if (tt != 0 ||tr[0][j] != '\0' ||(j >= 1 && tr[1][j - 1] != '\0') ||(j >= 2 && tr[2][j - 2] != '\0') ||(j >= 3 && tr[3][j - 3] != '\0') ||(j >= 4 && tr[4][j - 4] != '\0') ||(j >= 5 && tr[5][j - 5] != '\0'))rs[cn][j] = '0';else break;if (tr[0][j] != '\0') {rs[cn][j] = (tr[0][j] - '0' + tt) % 10 + '0';tt = (tr[0][j] - '0' + tt) / 10;}else {rs[cn][j] = tt % 10 + '0';tt = tt / 10;}if (j >= 1 && tr[1][j - 1] != '\0') {int tmp = rs[cn][j] - '0' + tr[1][j - 1] - '0';rs[cn][j] = tmp % 10 + '0';tt += tmp / 10;}if (j >= 2 && tr[2][j - 2] != '\0') {int tmp = rs[cn][j] - '0' + tr[2][j - 2] - '0';rs[cn][j] = tmp % 10 + '0';tt += tmp / 10;}if (j >= 3 && tr[3][j - 3] != '\0') {int tmp = rs[cn][j] - '0' + tr[3][j - 3] - '0';rs[cn][j] = tmp % 10 + '0';tt += tmp / 10;}if (j >= 4 && tr[4][j - 4] != '\0') {int tmp = rs[cn][j] - '0' + tr[4][j - 4] - '0';rs[cn][j] = tmp % 10 + '0';tt += tmp / 10;}if (j >= 5 && tr[5][j - 5] != '\0') {int tmp = rs[cn][j] - '0' + tr[5][j - 5] - '0';rs[cn][j] = tmp % 10 + '0';tt += tmp / 10;}}}for (int j = 0; rs[cn][j] != '\0'; j++) {if (rs[cn][j] != '0')break;else if (j <= ptn[cn])rs[cn][j] = '*';}bool flag = false;for (int j = 149; j >= 0; j--) {if (j - 1 == ptn[cn] && flag == false && rs[cn][j] == '0')continue;if (j == ptn[cn] && rs[cn][j - 1] != '\0'&&rs[cn][j - 1] != '*')cout << '.';if (rs[cn][j] != '\0'&&rs[cn][j] != '*') {cout << rs[cn][j];flag = true;}}cout << endl;cn++;}//system("pause");}

总结:1.对每个输入都要马上输出而不是最后一起输出

2.注意结果小于1去掉开头的0;去掉小数点后多余的0

3.注意指数等于0或1的情况

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