Sicily 1029. Rabbit

来源：互联网发布：男士护肤面膜知乎编辑：程序博客网时间：2024/06/03 16:47

题目要不是提醒了一下m = 2的时候发是fibonacci sequence以我的水平估计是看不出来数据模型~~~

所以还是愚蠢地用了列举方法，分别了m = 1, m = 2, m = 3然后d = 10的情况去观察规律。

最后还有个高精度加法，第一次接触，有点头疼，好多细节不完善，所以还是参考了下前辈的思路：点击打开链接

看完再自己动手写了一遍

好慢不知道主要是不是cin cout的原因，好想知道前面0.00sec的是怎么写的，期待告知！

// Problem#: 1029// Author: Reid Chan#include <iostream>#include <string>#include <cstring>using namespace std;void integer_2_string(int num, string &str){    str = "";    char ch;    while (num != 0) {        ch = (num % 10) + '0';        str = ch + str;        num /= 10;    }}string high_accuracy_addition(string left, string right){    int len_l = left.length();    int len_r = right.length();    while (len_l < len_r) {        left = '0' + left;        len_l++;    }    while (len_r < len_l) {        right = '0' + right;        len_r++;    }    left = '0' + left;    right = '0' + right;    int carry = 0;    string result = "";    char re;    int trans;    for (int i = left.length() - 1; i >= 0; --i) {        trans = (left[i] - '0' + right[i] - '0') + carry;        if (trans > 9) {            re = (trans - 10) + '0';            carry = 1;        } else {            re = trans + '0';            carry = 0;        }        result = re + result;    }    if (result[0] == '0') {        result = result.substr(1, result.length() - 1);    }    return result;}int main() {    int m, d;    string str[101];    while (cin >> m >> d && (m != 0 && d != 0)) {        int pre = 0, now = 0;        for (int i = 0; i < m; ++i) {            now = pre + 1;            integer_2_string(now, str[i]);            pre = now;        }        if (d < m) {            cout << str[d] << endl;        } else {            for (int i = m; i <= d; ++i) {                str[i] = high_accuracy_addition(str[i - 1], str[i - m]);            }            cout << str[d] << endl;        }    }    return 0;}

0 0