B1028 人口普查
某城镇进行人口普查,得到了全体居民的生日。现请你写个程序,找出镇上最年长和最年轻的人。这里确保每个输入的日期都是合法的,但不一定是合理的——假设已知镇上没有超过200岁的老人,而今天是2014年9月6日,所以超过200岁的生日和未出生的生日都是不合理的,应该被过滤掉。
输入格式
输入在第一行给出正整数N,取值在(0, 10^5];随后N行,每行给出1个人的姓名(由不超过5个英文字母组成的字符串)、以及按“yyyy/mm/dd”(即年/月/日)格式给出的生日。题目保证最年长和最年轻的人没有并列。
输出格式
在一行中顺序输出有效生日的个数、最年长人和最年轻人的姓名,其间以空格分隔。
输入样例
5
John 2001/05/12
Tom 1814/09/06
Ann 2121/01/30
James 1814/09/05
Steve 1967/11/20
输出样例
3 Tom John
idea
- 初始想法,把日期提取出来并比较
- 优化,把出生日期视为字符串,实际上年龄大小比较 等价于 字符串大小比较
注意不足两位数的月或日补零,保持对应位数比较
(柳神的想法让我感觉到了世界的参差┭┮﹏┭┮
solution1
#include <stdio.h>
#include <string.h>
struct resident{
char name[20];
int year;
int month;
int day;
};
int main(){
int n, count = 0;
resident find[2];
find[0].year = 2015;
find[1].year = 1813;
scanf("%d", &n);
resident r[n];
char time[15], t[10];
for(int i = 0; i< n; i++){
scanf("%s %s", r[i].name, time);
for(int j = 0; j < 4; j++)
t[j] = time[j];
sscanf(t, "%d", &r[i].year);
if(r[i].year <1814 || r[i].year > 2014)
continue;
memset(t, '\0', sizeof(t));
for(int j = 5, k = 0; j < 7; j++)
t[k++] = time[j];
sscanf(t, "%d", &r[i].month);
if((r[i].year == 2014 && r[i].month > 9) || (r[i].year == 1814 && r[i].month < 9))
continue;
for(int j = 8, k = 0; j < 10; j++)
t[k++] = time[j];
sscanf(t, "%d", &r[i].day);
if((r[i].year == 2014 && r[i].month == 9 && r[i].day > 6) || (r[i].year == 1814 && r[i].month == 9 && r[i].day < 6))
continue;
count++;
if(r[i].year < find[0].year || (r[i].year == find[0].year && r[i].month < find[0].month) ||(r[i].year == find[0].year &&r[i].month == find[0].month && r[i].day < find[0].day)){
strcpy(find[0].name, r[i].name);
find[0].year = r[i].year;
find[0].month = r[i].month;
find[0].day = r[i].day;
}
if(r[i].year > find[1].year || (r[i].year == find[1].year && r[i].month > find[1].month) ||(r[i].year == find[1].year &&r[i].month == find[1].month && r[i].day > find[1].day)){
strcpy(find[1].name, r[i].name);
find[1].year = r[i].year;
find[1].month = r[i].month;
find[1].day = r[i].day;
}
}
printf("%d", count);
if(find[0].year != 2015){
printf(" %s", find[0].name);
}
if(find[1].year != 1813){
printf(" %s", find[1].name);
}
return 0;
}
solution2
#include <stdio.h>
#include <string.h>
int main(){
int n, count = 0, y, m, d, yy[2] = {2015, 1813}, mm[2], dd[2];
char name[6], minName[6], maxName[6];
scanf("%d", &n);
while(n--){
scanf("%s %d/%d/%d", name, &y, &m, &d);
if(y > 2014 || (y == 2014 && m > 9) || (y == 2014 && m == 9 && d > 6) || y < 1814 ||(y == 1814 && m < 9) || (y == 1814 && m == 9 && d < 6))
continue;
count++;
if(y < yy[0] || (y == yy[0] && m < mm[0]) || (y == yy[0] && m == mm[0] && d < dd[0])){
yy[0] = y;
mm[0] = m;
dd[0] = d;
strcpy(maxName, name);
}
if(y > yy[1] || (y == yy[1] && m > mm[1]) || (y == yy[1] && m == mm[1] && d > dd[1])){
yy[1] = y;
mm[1] = m;
dd[1] = d;
strcpy(minName, name);
}
}
printf("%d", count);
if(strlen(maxName))
printf(" %s", maxName);
if(strlen(minName))
printf(" %s", minName);
return 0;
}
solution3
#include <stdio.h>
#include <string.h>
int main(){
int n, count = 0;
char birth[11], min[11] = "1814/09/05", max[11] = "2014/09/07", minName[6], maxName[6], name[6];
scanf("%d", &n);
while(n--){
scanf("%s %s", name, birth);
if(strcmp(birth, "2014/09/06") > 0 || strcmp(birth, "1814/09/06") < 0)
continue;
count++;
if(strcmp(birth, max) < 0){
strcpy(max, birth);
strcpy(maxName, name);
}
if(strcmp(birth, min) > 0){
strcpy(min, birth);
strcpy(minName, name);
}
}
printf("%d", count);
if(strlen(maxName))
printf(" %s", maxName);
if(strlen(minName))
printf(" %s", minName);
return 0;
}