2、顺子日期
题目 顺子日期
思路分析
从2022.1.1枚举到2023
看每个日期是否符合三连顺
代码实现
#include <bits/stdc++.h>
using namespace std;
int days[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
bool isleap(int y){
return y%100 && y%4==0 || y%400==0;
}
int getdays(int y,int m){
return days[m]+(m==2 && isleap(y));
}
void nextday(int &y,int &m,int &d){
d++;
if(d>getdays(y,m)){
d=1;
m++;
if(m>12){
m=1;
y++;
}
}
}
bool check(char* s){
for(int i=2;i<8;i++){
if(s[i-1]-s[i-2]==1 && s[i]-s[i-1]==1)
return true;
}
return false;
}
int main()
{
int cury=2022,curm=1,curd=1;
int cnt=0;
while(cury<2023){
char s[10];
sprintf(s,"%04d%02d%02d",cury,curm,curd);
if(check(s))
cnt++;
nextday(cury,curm,curd);
}
cout<<cnt;
return 0;
}
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
//#define int long long 蓝桥杯用不了signed main
using ll = long long;
using ull = unsigned long long;
using PII = pair<int,int>;
using Pll = pair<ll,ll>;
int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
const int inf = 0x3f3f3f3f;
int days[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
bool is_leap(int y){
return y%100 && y%4==0 || y%400==0;
}
int get_days(int y,int m){
return days[m]+(m==2 && is_leap(y));
}
void next_day(int &y,int &m,int &d){
d++;
if(d>get_days(y,m)){
d=1;
m++;
if(m>12){
m=1;
y++;
}
}
}
bool check_date(int y,int m,int d){
if(m<1 || m>12) return false;
if(d<1 || d>get_days(y,m)) return false;
return true;
}
/*
while((curYear < targetYear) ||
(curYear == targetYear && curMonth < targetMonth) ||
(curYear == targetYear && curMonth == targetMonth && curDay < targetDay)){
next_day();
}
*/
bool check(string s){
for(int i=2;i<s.size();i++){
if(s[i-1]-s[i-2]==1 && s[i]-s[i-1]==1) return true;
}
return false;
}
int main()
{
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int curYear=2022,curMonth=1,curDay=1;
int cnt=0;
while(curYear<2023){
char buffer[9];
sprintf(buffer, "%04d%02d%02d", curYear, curMonth, curDay);
string s(buffer);
if(check(s)) cnt++;
next_day(curYear,curMonth,curDay);
}
cout<<cnt;
return 0;
}
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