1028 人口普查（附详细注释，逻辑分析）

写在前面

• 实现思路

• 结构体封装姓名、出生年月人的元数据信息
• 巧妙利用scanf读取格式
• 出生日期有效性判断

• 算法笔记代码略冗长，熟练的话10分钟内a题

测试用例

input:
5
John 2001/05/12
Tom 1814/09/06
Ann 2121/01/30
James 1814/09/05
Steve 1967/11/20

output:
3

ac代码

#include<cstdio>
struct person
{
    char name[10];
    int yy, mm, dd;
} oldest, youngest, left, right, tmp;
bool LessEqu(person a, person b)
{
    if(a.yy != b.yy) return a.yy <= b.yy;
    else if(a.mm != b.mm) return a.mm <= b.mm;
    else return a.dd <= b.dd;
}
bool MoreEqu(person a, person b)
{
    if(a.yy != b.yy) return a.yy >= b.yy;
    else if(a.mm != b.mm) return a.mm >= b.mm;
    else return a.dd >= b.dd;
}
void init()
{
    youngest.yy = left.yy = 1814;
    oldest.yy = right.yy = 2014;
    youngest.mm = oldest.mm = left.mm = right.mm = 9;
    youngest.dd = oldest.dd = left.dd = right.dd = 6;
}

int main()
{
    init();
    int n, num=0;
    scanf("%d", &n);

    for(int i=0; i<n; i++)
    {
        scanf("%s %d/%d/%d", tmp.name,&tmp.yy,&tmp.mm,&tmp.dd);
        if(MoreEqu(tmp, left) && LessEqu(tmp, right))
        {
            num++;
            if(LessEqu(tmp, oldest)) oldest = tmp;
            if(MoreEqu(tmp, youngest)) youngest = tmp;
        }
    }
    if(num==0) printf("0\n");
    else printf("%d %s %s", num, oldest.name, youngest.name);

    return 0;
}

参考代码

• ​​1028. 人口普查(20).cpp​​

#include <iostream>
using namespace std;
int main() {
    int n, cnt = 0;
    cin >> n;
    string name, birth, maxname, minname, maxbirth = "1814/09/06", minbirth = "2014/09/06";
    for (int i = 0; i < n; i++) {
        cin >> name >> birth;
        if (birth >= "1814/09/06" && birth <= "2014/09/06") {
            cnt++;
            if (birth >= maxbirth) {
                maxbirth = birth;
                maxname = name;
            }
            if (birth <= minbirth) {
                minbirth = birth;
                minname = name;
            }
        }
    }
    cout << cnt;
    if (cnt != 0) cout << " " << minname << " " << maxname;
    return 0;
}

C++ 字符串比较

• ​​C++ 字符串可以和类型相同的字符串相比较，也可以和具有同样字符类型的数组比较​​

• 串值相同，返回 0；若字符串 S 字典序先于 S2，返回负值；反之，返回正值

// m = 1, n = -1
string A ("aBcdef");
string B ("AbcdEf");
int m=A.compare (B); //完整的A和B的比较
int n=A.compare(1,5,B,4,2); //"Bcdef"和"AbcdEf"比较