54 lines
1.8 KiB
C++
54 lines
1.8 KiB
C++
// https://www.luogu.com.cn/problem/P2010
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#include <bits/stdc++.h>
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using namespace std;
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//月份日期常数数组
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int months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
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//判断是不是闰年
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bool LeapYear(int year) {
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return ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0);
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}
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//是不是回文字符串
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bool isHW(string s) {
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string t = s;
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reverse(t.begin(), t.end());
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return s == t;
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}
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//通过年月日的整数类型值，构建出一个字符串的年月日
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//数字拼接出字符串办法
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string build(int year, int month, int date) {
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char x[10]; //不太会使用这个char数组，因为最后一个位置需要放置\0,所以需要开大一点点
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sprintf(x, "%04d%02d%02d", year, month, date); //拼接出一个字符串
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return x; //将字符数组转为字符串，直接赋值即可
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}
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int cnt;
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int main() {
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string st, ed;
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cin >> st >> ed;
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int y1 = stoi(st.substr(0, 4)); //截取字符串，stoi:字符串转数字
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int y2 = stoi(ed.substr(0, 4));
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//枚举年份
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for (int i = y1; i <= y2; i++)
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// 枚举月份
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for (int j = 1; j <= 12; j++) {
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//枚举日期
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int day = months[j];
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if (LeapYear(i) && j == 2) day = 29; //闰年二月份，共29天
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for (int k = 1; k <= day; k++) {
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//合法日期，构建成字符串
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string s = build(i, j, k);
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//与范围字符串st,ed进行比较，如果不在范围内，continue
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if (s < st || s > ed) continue; //日期字符串符合字典序
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//判断是不是回文
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if (isHW(s)) cnt++;
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}
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}
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//输出
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printf("%d\n", cnt);
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return 0;
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} |