日期题:周一或月初+2 其它+1问题
题目描述
本题为填空题,只需要算出结果后,在代码中使用输出语句将所填结果输出即可。
小蓝每天都锻炼身体。
正常情况下,小蓝每天跑 1 千米。如果某天是周一或者月初(1 日),为了激励自己,小蓝要跑 2 千米。如果同时是周一或月初,小蓝也是跑 2 千米。
小蓝跑步已经坚持了很长时间,从 2000 年 1 月 1 日周六(含)到 2020 年 10 月 1 日周四(含)。请问这段时间小蓝总共跑步多少千米?
运行限制
- 最大运行时间:1s
- 最大运行内存: 128M
代码:
1 #include <iostream> 2 using namespace std; 3 4 int mrun[] = {0, 31,29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; 5 int mping[] = {0,31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; 6 7 bool run (int n ) 8 { 9 if( n % 4 == 0 && n % 100 != 0 || n % 400 == 0) 10 { 11 return true; 12 } 13 else 14 { 15 return false; 16 } 17 18 } 19 20 int main() 21 { 22 int week = 5; //6 23 24 int sum = 0; 25 26 for(int y = 2000; y <= 2020; y ++) 27 { 28 29 for(int m = 1; m <= 12; m ++) 30 { 31 if(run(y)) 32 { 33 for(int d = 1; d <= mrun[m]; d++) 34 { 35 if(week == 0 || d == 1){ 36 sum += 2; 37 } 38 else 39 { 40 sum += 1; 41 } 42 week = (week + 1 ) % 7; 43 44 if(y == 2020 && m == 10 && d == 1) 45 { 46 cout << sum << endl; 47 } 48 } 49 } 50 else 51 { 52 for(int d = 1; d<= mping[m]; d++) 53 { 54 if(d== 1 || week == 0) 55 { 56 sum += 2; 57 } 58 else 59 { 60 sum += 1; 61 } 62 week = (week +1 ) % 7; 63 64 if(y == 2020 && m == 10 && d == 1) 65 { 66 cout << sum << endl; 67 } 68 } 69 70 } 71 72 } 73 74 } 75 76 77 return 0; 78 }
#include <iostream>
using namespace std;
int mrun[] = {0, 31,29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int mping[] = {0,31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool run (int n )
{
if( n % 4 == 0 && n % 100 != 0 || n % 400 == 0)
{
return true;
}
else
{
return false;
}
}
int main()
{
int week = 5; //6
int sum = 0;
for(int y = 2000; y <= 2020; y ++)
{
for(int m = 1; m <= 12; m ++)
{
if(run(y))
{
for(int d = 1; d <= mrun[m]; d++)
{
if(week == 0 || d == 1){
sum += 2;
}
else
{
sum += 1;
}
week = (week + 1 ) % 7;
if(y == 2020 && m == 10 && d == 1)
{
cout << sum << endl;
}
}
}
else
{
for(int d = 1; d<= mping[m]; d++)
{
if(d== 1 || week == 0)
{
sum += 2;
}
else
{
sum += 1;
}
week = (week +1 ) % 7;
if(y == 2020 && m == 10 && d == 1)
{
cout << sum << endl;
}
}
}
}
}
return 0;
}

浙公网安备 33010602011771号