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poj 2247 Humble Numbers
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Humble Numbers
| Time Limit:?1000MS | ? | Memory Limit:?65536K |
| Total Submissions:?9453 | ? | Accepted:?4440 |
題目大意：找出全部因子中僅僅有2,3,5,7的數(shù)。給出n，求出第n個這樣 ? 僅僅有6000不到打表
注意第11 ?12 ?13 的輸出與 1 ? 2 ? 3 ? ?幾十一 ? 幾十二 ? 幾十三 的輸出不同 ?
#include <iostream>
#include <algorithm>
using namespace std;long humble[6000];int main()
{int i, n2, n3, n5, n7;humble[1] = 1;n2 = 1;n3 = 1;n5 = 1;n7 = 1;for (i = 2; i <= 5900; i++){humble[i] = min(min(humble[n2]*2, humble[n3]*3), min(humble[n5]*5, humble[n7]*7));if (humble[i] == humble[n2]*2)n2++;if (humble[i] == humble[n3]*3)n3++;if (humble[i] == humble[n5]*5)n5++;if (humble[i] == humble[n7]*7)n7++;}int num;while (cin >> num && num){int tmp1, tmp2;tmp1 = num % 10;tmp2 = num % 100;if (tmp1 == 1){if (tmp2 == 11)cout << "The " << num << "th humble number is " << humble[num] << "." << endl;else cout << "The " << num << "st humble number is " << humble[num] << "." << endl;}else if (tmp1 == 2){if (tmp2 == 12)cout << "The " << num << "th humble number is " << humble[num] << "." << endl;elsecout << "The " << num << "nd humble number is " << humble[num] << "." << endl;}else if (tmp1 == 3){if (tmp2 == 13)cout << "The " << num << "th humble number is " << humble[num] << "." << endl;elsecout << "The " << num << "rd humble number is " << humble[num] << "." << endl;}else cout << "The " << num << "th humble number is " << humble[num] << "." << endl;}system("pause");
}
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