文章目錄 1. REPEAT程序 2. 猜生日 3. 成績統計 4. 乘積尾零 5. 等差數列 6. 合并檢測 7. 回文日期 8. 紀念日 9. 既約分數 10. 門牌制作 11. 年號字符 12. 跑步鍛煉 13. 跑步訓練 14. 平方和 15. 七段數碼管 16. 奇數倍數 17. 刪除字符 18. 蛇形填數 19. 數的分解 20. 數列求值 21. 特別數的和 22. 完全二叉樹的權值 23. 整數序列 24. 質數 25. 走方格 26. 排列字母 27. 日期問題 28. 飲料換購 29. 受傷的皇后 30. 最大降雨量 31. 單詞分析 32. 卡片 33. 空間 34. 時間顯示 35. 分巧克力
1. REPEAT程序
題目描述:
REPEAT
2 ; A
= A
+ 4 ; REPEAT
5 ; REPEAT
6 ; A
= A
+ 5 ; A
= A
+ 7 ; A
= A
+ 8 ;
A
= A
+ 9 ;
該片段中從 A = A + 4 所在的行到 A = A + 8 所在的行都在第一行的循環兩次中。
題目分析:
# include <iostream> using namespace std
; int main ( ) { int A
= 0 ; for ( int i
= 0 ; i
< 2 ; i
++ ) { A
+= 4 ; for ( int j
= 0 ; j
< 5 ; j
++ ) { for ( int k
= 0 ; k
< 6 ; k
++ ) { A
+= 5 ; } A
+= 7 ; } A
+= 8 ; } A
+= 9 ; cout
<< "A = " << A
<< endl
;
}
2. 猜生日
題目描述：
# include <iostream> using namespace std
;
int main ( )
{ for ( int i
= 19000101 ; i
<= 20120312 ; i
++ ) { if ( i
% 2012 == 0 && i
% 3 == 0 && i
% 12 == 0 && i
/ 100 % 100 == 6 && i
% 100 < 32 ) { cout
<< i
; break ; } } return 0 ;
}
3. 成績統計
題目描述:
【輸入格式】 輸入的第一行包含一個整數n，表示考試人數。接下來n行，每行包含一個0至100的整數，表示一個學生的得分。
# include <iostream> using namespace std
; int main ( ) { int n
, bad
= 0 , good
= 0 , b
, f
; cin
>> n
; int a
[ n
] ; for ( int i
= 0 ; i
< n
; i
++ ) { cin
>> a
[ i
] ; if ( a
[ i
] >= 85 ) { good
++ ; } if ( a
[ i
] >= 60 ) { bad
++ ; } } b
= bad
* 100.0 / n
* 1.0 + 0.5 ; f
= good
* 100.0 / n
* 1.0 + 0.5 ; cout
<< b
<< "%" << endl
; cout
<< f
<< "%" << endl
;
}
4. 乘積尾零
題目描述：
5650 4542 3554 473 946 4114 3871 9073 90 4329
2758 7949 6113 5659 5245 7432 3051 4434 6704 3594
9937 1173 6866 3397 4759 7557 3070 2287 1453 9899
1486 5722 3135 1170 4014 5510 5120 729 2880 9019
2049 698 4582 4346 4427 646 9742 7340 1230 7683
5693 7015 6887 7381 4172 4341 2909 2027 7355 5649
6701 6645 1671 5978 2704 9926 295 3125 3878 6785
2066 4247 4800 1578 6652 4616 1113 6205 3264 2915
3966 5291 2904 1285 2193 1428 2265 8730 9436 7074
689 5510 8243 6114 337 4096 8199 7313 3685 211
# include <iostream> # include <cmath> using namespace std
;
int main ( )
{ int a
[ 100 ] = { 5650 , 4542 , 3554 , 473 , 946 , 4114 , 3871 , 9073 , 90 , 4329 , 2758 , 7949 , 6113 , 5659 , 5245 , 7432 , 3051 , 4434 , 6704 , 3594 , 9937 , 1173 , 6866 , 3397 , 4759 , 7557 , 3070 , 2287 , 1453 , 9899 , 1486 , 5722 , 3135 , 1170 , 4014 , 5510 , 5120 , 729 , 2880 , 9019 , 2049 , 698 , 4582 , 4346 , 4427 , 646 , 9742 , 7340 , 1230 , 7683 , 5693 , 7015 , 6887 , 7381 , 4172 , 4341 , 2909 , 2027 , 7355 , 5649 , 6701 , 6645 , 1671 , 5978 , 2704 , 9926 , 295 , 3125 , 3878 , 6785 , 2066 , 4247 , 4800 , 1578 , 6652 , 4616 , 1113 , 6205 , 3264 , 2915 , 3966 , 5291 , 2904 , 1285 , 2193 , 1428 , 2265 , 8730 , 9436 , 7074 , 689 , 5510 , 8243 , 6114 , 337 , 4096 , 8199 , 7313 , 3685 , 211 } ; int cnt2
= 0 , cnt5
= 0 ; for ( int i
= 0 ; i
<= 99 ; i
++ ) { int t
= a
[ i
] ; while ( t
% 2 == 0 ) { cnt2
++ ; t
/= 2 ; } while ( t
% 5 == 0 ) { cnt5
++ ; t
/= 5 ; } } cout
<< min ( cnt2
, cnt5
) << endl
; return 0 ;
}
5. 等差數列
題目描述：
【輸入描述】
# include <iostream> # include <algorithm> using namespace std
;
int main ( )
{ int n
, min
= 10000 ; cin
>> n
; long long a
[ n
] ; for ( int i
= 0 ; i
< n
; i
++ ) { cin
>> a
[ i
] ; } sort ( a
, a
+ n
) ; for ( int j
= 0 ; j
< n
- 1 ; j
++ ) { if ( a
[ j
+ 1 ] - a
[ j
] < min
) { min
= a
[ j
+ 1 ] - a
[ j
] ; } } int m
= ( a
[ n
- 1 ] - a
[ 0 ] ) / min
; if ( min
== 0 ) cout
<< n
; else cout
<< m
+ 1 ; return 0 ;
}
6. 合并檢測
題目描述:
A 國估計被測的民眾的感染率大概是 1，呈均勻分布。請問 k 取多少能最節省試劑盒？
題目分析:
# include <iostream> using namespace std
; int main ( ) { int k
, sum
, j
, s
= 1000 , min
, m
, f
; for ( k
= 1 ; k
<= 100 ; k
++ ) { if ( 100 % k
== 0 ) { j
= 100 / k
; sum
= j
+ k
; if ( sum
< s
) { min
= sum
; s
= sum
; f
= k
; } } else { j
= 100 / k
; sum
= j
+ 1 + k
; if ( sum
< s
) { min
= sum
; s
= sum
; f
= k
; } } } cout
<< "最少需要試劑 " << min
<< " 盒，此時以 " << f
<< " 人為一個采集標本" ;
}
7. 回文日期
題目描述：
【輸入格式】
【輸出格式】
# include <iostream> using namespace std
; int leap ( int year
) { if ( ( year
% 4 == 0 && year
% 100 != 0 ) || year
% 400 == 0 ) return 1 ;
} int palindromic ( int num
) { int m
, m1
, m2
, m3
, m4
, m5
, m6
, m7
, m8
; m1
= ( num
% 10 ) * 10000000 ; m2
= ( num
/ 10 % 10 ) * 1000000 ; m3
= ( num
/ 100 % 10 ) * 100000 ; m4
= ( num
/ 1000 % 10 ) * 10000 ; m5
= ( num
/ 10000 % 10 ) * 1000 ; m6
= ( num
/ 100000 % 10 ) * 100 ; m7
= ( num
/ 1000000 % 10 ) * 10 ; m8
= num
/ 10000000 ; m
= m1
+ m2
+ m3
+ m4
+ m5
+ m6
+ m7
+ m8
; if ( num
== m
) return 1 ; else return 0 ;
} int main ( ) { int monthes
[ 13 ] = { 0 , 31 , 28 , 31 , 30 , 31 , 30 , 31 , 31 , 30 , 31 , 30 , 31 } ; int year
, month
, day
, n
, year0
, month0
, day0
, n1
, n2
, n3
, n4
, n5
, n6
, n7
, n8
, m
, m1
, m2
, m3
, m4
, m5
, m6
, m7
, m8
, k1
, k2
, k3
, k4
, k5
, k6
, k7
, k8
; int num
; cin
>> num
; if ( ! palindromic ( num
) ) return 0 ; n1
= ( num
/ 10000000 ) * 1000 ; n2
= ( num
/ 1000000 % 10 ) * 100 ; n3
= ( num
/ 100000 % 10 ) * 10 ; n4
= ( num
/ 10000 % 10 ) ; n5
= ( num
/ 1000 % 10 ) * 10 ; n6
= ( num
/ 100 % 10 ) ; n7
= ( num
/ 10 % 10 ) * 10 ; n8
= ( num
% 10 ) ; year0
= n1
+ n2
+ n3
+ n4
; month0
= n5
+ n6
; day0
= n7
+ n8
; for ( year
= year0
; year
< 10000 ; year
++ ) { if ( leap ( year
) == 1 ) monthes
[ 2 ] = 29 ; else monthes
[ 2 ] = 28 ; for ( month
= month0
; month
< 13 ; month
++ ) { for ( day
= day0
+ 1 ; day
< monthes
[ month
] ; day
++ ) { n
= year
* 10000 + month
* 100 + day
; m1
= ( day
% 10 ) * 10000000 ; m2
= ( day
/ 10 ) * 1000000 ; m3
= ( month
% 10 ) * 100000 ; m4
= ( month
/ 10 ) * 10000 ; m5
= ( year
% 10 ) * 1000 ; m6
= ( year
/ 10 % 10 ) * 100 ; m7
= ( year
/ 100 % 10 ) * 10 ; m8
= year
/ 1000 ; k1
= ( day
% 10 ) ; k2
= ( day
/ 10 ) ; k3
= ( month
% 10 ) ; k4
= ( month
/ 10 ) ; k5
= ( year
% 10 ) ; k6
= ( year
/ 10 % 10 ) ; k7
= ( year
/ 100 % 10 ) ; k8
= year
/ 1000 ; m
= m1
+ m2
+ m3
+ m4
+ m5
+ m6
+ m7
+ m8
; if ( ( n
== m
) && ( k8
== k1
) && ( k6
== k1
) && ( k3
== k1
) && ( k7
== k2
) && ( k5
== k2
) && ( k4
== k2
) ) { cout
<< n
<< endl
; return 0 ; } } } }
}
8. 紀念日
題目描述:
題目分析:
# include <iostream> using namespace std
; int main ( ) { int i
, j
, flag
, sum
= 0 ; for ( i
= 1922 ; i
<= 2020 ; i
++ ) { if ( ( i
% 4 == 0 && i
% 100 != 0 ) || i
% 400 == 0 ) { sum
+= 366 ; } else sum
+= 365 ; } sum
-= 22 ; cout
<< sum
* 24 * 60 ;
}
9. 既約分數
題目描述:
題目分析:
# include <iostream> using namespace std
; int panduan ( int a
, int b
)
{ int c
; if ( a
== 1 && b
== 1 ) return 1 ; if ( a
>= b
) { while ( c
= a
% b
) { a
= b
; b
= c
; } if ( b
== 1 ) return 1 ; else return 0 ; } else { while ( c
= b
% a
) { b
= a
; a
= c
; } if ( a
== 1 ) return 1 ; else return 0 ; }
}
int main ( ) { int sum
= 0 ; for ( int i
= 1 ; i
<= 2020 ; i
++ ) { for ( int j
= 1 ; j
<= 2020 ; j
++ ) { sum
+= panduan ( i
, j
) ; } } cout
<< sum
<< endl
; cout
<< panduan ( 2 , 21 ) ;
}
10. 門牌制作
題目描述:
# include <iostream> using namespace std
; int panduan ( int num
) { int sum
= 0 ; while ( num
) { if ( num
% 10 == 2 ) { sum
++ ; } num
= num
/ 10 ; } return sum
;
} int main ( ) { int n
; for ( int i
= 0 ; i
<= 2020 ; i
++ ) { n
+= panduan ( i
) ; } cout
<< n
;
}
11. 年號字符
題目描述:
題目分析:
# include <iostream> # include <stack> using namespace std
; int main ( ) { int n
= 2019 ; stack
< int > s
; int temp
; while ( n
) { temp
= n
% 26 ; s
. push ( temp
) ; n
/= 26 ; } while ( ! s
. empty ( ) ) { char c
= s
. top ( ) + 64 ; cout
<< c
; s
. pop ( ) ; }
}
12. 跑步鍛煉
題目描述:
題目分析:
# include <iostream> using namespace std
; int panduan ( int a
) { if ( ( a
% 4 == 0 && a
% 100 != 0 ) || a
% 400 == 0 ) return 1 ; else return 0 ;
} int main ( ) { int sum
= 0 ; int monthes
[ 13 ] = { 0 , 31 , 28 , 31 , 30 , 31 , 30 , 31 , 31 , 30 , 31 , 30 , 31 } ; int mon
= 6 ; for ( int year
= 2000 ; year
<= 2020 ; year
++ ) { if ( panduan ( year
) ) monthes
[ 2 ] = 29 ; else monthes
[ 2 ] = 28 ; for ( int month
= 1 ; month
<= 12 ; month
++ ) { for ( int day
= 1 ; day
<= monthes
[ month
] ; day
++ ) { if ( mon
== 8 ) mon
= 1 ; if ( mon
== 1 && day
== 1 ) { sum
+= 2 ; } else if ( mon
== 1 || day
== 1 ) { sum
+= 2 ; } else sum
+= 1 ; if ( year
== 2020 && month
== 10 && day
== 1 ) { cout
<< sum
; return 0 ; } mon
++ ; } } }
}
13. 跑步訓練
題目描述:
題目分析:
# include <iostream> using namespace std
; int main ( ) { int sum
= 0 ; int N
= 10000 ; while ( N
> 0 ) { N
-= 600 ; sum
+= 60 ; N
+= 300 ; sum
+= 60 ; if ( N
<= 600 ) { sum
+= ( N
/ 10 ) ; break ; } } cout
<< sum
;
}
14. 平方和
題目描述：
# include <iostream> # include <string> using namespace std
;
int main ( )
{ long long sum
= 0 ; for ( int i
= 1 ; i
<= 2019 ; i
++ ) { string str
= to_string ( i
) ; if ( str
. find ( "2" ) == - 1 && str
. find ( "0" ) == - 1 && str
. find ( "1" ) == - 1 && str
. find ( "9" ) == - 1 ) continue ; else sum
+= i
* i
; } cout
<< sum
; return 0 ;
}
15. 七段數碼管
題目描述:
思路：
# include <iostream> using namespace std
; int main ( )
{ int sum
= 0 ; for ( int a
= 0 ; a
<= 1 ; a
++ ) { for ( int b
= 0 ; b
<= 1 ; b
++ ) { for ( int c
= 0 ; c
<= 1 ; c
++ ) { for ( int d
= 0 ; d
<= 1 ; d
++ ) { for ( int e
= 0 ; e
<= 1 ; e
++ ) { for ( int f
= 0 ; f
<= 1 ; f
++ ) { for ( int g
= 0 ; g
<= 1 ; g
++ ) { int s
= a
+ b
+ c
+ d
+ e
+ f
+ g
; if ( s
== 1 ) { sum
++ ; } else if ( s
>= 2 ) { if ( a
== 1 && b
== 0 && f
== 0 ) continue ; else if ( b
== 1 && a
== 0 && c
== 0 && g
== 0 ) continue ; else if ( c
== 1 && b
== 0 && g
== 0 && d
== 0 ) continue ; else if ( d
== 1 && e
== 0 && c
== 0 ) continue ; else if ( e
== 1 && d
== 0 && g
== 0 && f
== 0 ) continue ; else if ( f
== 1 && a
== 0 && e
== 0 && g
== 0 ) continue ; sum
++ ; } } } } } } } } printf ( "%d\n" , sum
- 3 ) ; return 0 ;
}
16. 奇數倍數
題目描述：
# include <iostream> using namespace std
; int main ( )
{ int n1
, n2
, n3
, n4
, n5
, n6
; for ( int i
= 2019 ; ; i
++ ) { n1
= i
% 10 ; n2
= i
/ 10 % 10 ; n3
= i
/ 100 % 10 ; n4
= i
/ 1000 % 10 ; n5
= i
/ 10000 % 10 ; n6
= i
/ 100000 % 10 ; if ( i
% 2019 == 0 && n1
% 2 == 1 && n2
% 2 == 1 && n3
% 2 == 1 && n4
% 2 == 1 && n5
% 2 == 1 && n6
% 2 == 1 ) { cout
<< i
; return 0 ; } } return 0 ;
}
17. 刪除字符
題目描述：
【輸入描述】
# include <iostream> # include <string> using namespace std
;
int main ( )
{ string str
; int n
; cin
>> str
; cin
>> n
; int f
= str
. length ( ) - 1 ; while ( n
-- ) { for ( int i
= 0 ; i
< str
. length ( ) - 1 ; i
++ ) { if ( str
[ i
] > str
[ i
+ 1 ] ) { f
= i
; break ; } } str
[ f
] = '0' ; } for ( int j
= 0 ; j
< str
. length ( ) ; j
++ ) { if ( str
[ j
] != '0' ) cout
<< str
[ j
] ; } return 0 ;
}
18. 蛇形填數
題目描述:
題目分析:
1 行
1 列
1 3 2 + 4
2 行
2 列
4 5 6 10 9 8 7 + 8
3 行
3 列
11 12 13 14 15 16 17 18 19 20 21 + 12
4 行
4 列
22 23 24 25 26 27 28
# include <iostream> using namespace std
; int main ( ) { int m
= 1 ; for ( int i
= 0 ; i
< 20 ; i
++ ) { m
+= 4 * i
; } cout
<< m
;
}
19. 數的分解
題目描述：
# include <iostream> using namespace std
; bool check ( int n
) { for ( int i
= 0 ; n
!= 0 ; i
++ ) { if ( n
% 10 == 2 || n
% 10 == 4 ) return false ; else n
/= 10 ; } return true ;
} int main ( )
{ int num
= 0 ; for ( int num1
= 1 ; num1
<= 2019 / 3 ; num1
++ ) { if ( check ( num1
) ) { for ( int num2
= num1
+ 1 ; num2
<= ( 2019 - num1
) / 2 ; num2
++ ) { if ( check ( num2
) ) { int num3
= 2019 - num1
- num2
; if ( check ( num3
) && num3
> num2
) num
++ ; } } } } cout
<< num
; return 0 ;
}
20. 數列求值
題目描述:
題目分析:
# include <iostream> using namespace std
; int main ( ) { int n0
= 0 , n1
= 1 , n2
= 1 , n3
= 1 , n
= 3 ; long long int s
= 0 ; while ( n
) { s
= n1
+ n2
+ n3
; s
%= 10000 ; n1
= n2
; n2
= n3
; n3
= s
; n
++ ; if ( n
== 20190324 ) break ; } cout
<< s
;
}
21. 特別數的和
題目描述：
【輸入格式】
# include <iostream> using namespace std
; int main ( ) { int n0
, n1
, n2
, n3
, n4
, sum
= 0 , num
= 0 , N
, Num
= 0 ; cin
>> N
; n0
= N
; while ( n0
) { Num
++ ; n0
/= 10 ; } for ( int i
= 1 ; i
<= N
; i
++ ) { if ( Num
== 4 ) { if ( i
< 10 ) { n1
= i
% 10 ; if ( n1
== 0 || n1
== 1 || n1
== 2 || n1
== 9 ) { sum
+= i
; num
++ ; } } else if ( i
>= 10 && i
< 100 ) { n1
= i
/ 10 % 10 ; n2
= i
% 10 ; if ( ( n1
== 0 || n1
== 1 || n1
== 2 || n1
== 9 ) || ( n2
== 0 || n2
== 1 || n2
== 2 || n2
== 9 ) ) { sum
+= i
; num
++ ; } } else if ( i
>= 100 && i
< 1000 ) { n1
= i
/ 100 % 10 ; n2
= i
/ 10 % 10 ; n3
= i
% 10 ; if ( ( n1
== 2 || n1
== 0 || n1
== 1 || n1
== 9 ) || ( n2
== 2 || n2
== 0 || n2
== 1 || n2
== 9 ) || ( n3
== 2 || n3
== 0 || n3
== 1 || n3
== 9 ) ) { sum
+= i
; num
++ ; } } else { n1
= i
/ 1000 % 10 ; n2
= i
/ 100 % 10 ; n3
= i
/ 10 % 10 ; n4
= i
% 10 ; if ( ( n1
== 2 || n1
== 0 || n1
== 1 || n1
== 9 ) || ( n2
== 2 || n2
== 0 || n2
== 1 || n2
== 9 ) || ( n3
== 2 || n3
== 0 || n3
== 1 || n3
== 9 ) || ( n4
== 2 || n4
== 0 || n4
== 1 || n4
== 9 ) ) { sum
+= i
; num
++ ; } } } else if ( Num
== 3 ) { if ( i
< 10 ) { n1
= i
% 10 ; if ( n1
== 0 || n1
== 1 || n1
== 2 || n1
== 9 ) { sum
+= i
; num
++ ; } } else if ( i
>= 10 && i
< 100 ) { n1
= i
/ 10 % 10 ; n2
= i
% 10 ; if ( ( n1
== 0 || n1
== 1 || n1
== 2 || n1
== 9 ) || ( n2
== 0 || n2
== 1 || n2
== 2 || n2
== 9 ) ) { sum
+= i
; num
++ ; } } else { n1
= i
/ 100 % 10 ; n2
= i
/ 10 % 10 ; n3
= i
% 10 ; if ( ( n1
== 2 || n1
== 0 || n1
== 1 || n1
== 9 ) || ( n2
== 2 || n2
== 0 || n2
== 1 || n2
== 9 ) || ( n3
== 2 || n3
== 0 || n3
== 1 || n3
== 9 ) ) { sum
+= i
; num
++ ; } } } else if ( Num
== 2 ) { if ( i
< 10 ) { n1
= i
% 10 ; if ( n1
== 0 || n1
== 1 || n1
== 2 || n1
== 9 ) { sum
+= i
; num
++ ; } } else { n1
= i
/ 10 % 10 ; n2
= i
% 10 ; if ( ( n1
== 0 || n1
== 1 || n1
== 2 || n1
== 9 ) || ( n2
== 0 || n2
== 1 || n2
== 2 || n2
== 9 ) ) { sum
+= i
; num
++ ; } } } else if ( Num
== 1 ) { n1
= i
% 10 ; if ( ( n1
== 2 || n1
== 0 || n1
== 1 || n1
== 9 ) ) { sum
+= i
; num
++ ; } } } cout
<< sum
<< endl
;
}
22. 完全二叉樹的權值
題目描述:
【輸入格式】
# include <iostream> # include <cmath> using namespace std
; int main ( ) { int depthes
[ 1000 ] = { 0 } , N
, depth
, sd
[ 100 ] = { 0 } , max
= 0 , f
, h
; cin
>> N
; depth
= ceil ( log2 ( N
+ 1 ) ) ; for ( int i
= 1 ; i
<= N
; i
++ ) { cin
>> depthes
[ i
] ; } for ( int j
= 1 ; j
<= N
; j
++ ) { h
= ceil ( log2 ( j
+ 1 ) ) ; sd
[ h
] += depthes
[ j
] ; } for ( int k
= 1 ; k
<= depth
; k
++ ) { if ( sd
[ k
] > max
) { max
= sd
[ k
] ; f
= k
; } } cout
<< f
<< endl
;
}
23. 整數序列
題目描述:
【輸入格式】
【評測用例規模與約定】
# include <iostream> # include <string.h> using namespace std
; int main ( ) { int a
[ 100 ] , j
= 0 , i
, s
; cin
>> i
; while ( i
) { a
[ j
] = i
/ 2 ; i
= i
/ 2 ; j
++ ; s
++ ; } for ( int m
= 0 ; m
< s
- 2 ; m
++ ) { cout
<< a
[ m
] ; if ( a
[ m
+ 1 ] != 0 ) cout
<< "," ; }
}
24. 質數
題目描述：
# include <iostream> using namespace std
;
int main ( )
{ int flag
= 0 ; int num
= 2 ; for ( int i
= 5 ; i
< 20000 ; i
++ ) { flag
= 0 ; for ( int j
= 2 ; j
< i
; j
++ ) { if ( i
% j
== 0 ) { flag
= 1 ; break ; } } if ( flag
== 0 ) { num
++ ; if ( num
== 2019 ) cout
<< i
; } } return 0 ;
}
25. 走方格
題目描述:
【輸入格式】
【輸出格式】
【評測用例規模與約定】
# include <iostream> using namespace std
; int main ( ) { int n
, m
, i
, j
; int a
[ 35 ] [ 35 ] ; cin
>> n
>> m
; for ( i
= 0 ; i
< n
; i
++ ) a
[ i
] [ 0 ] = 1 ; for ( j
= 0 ; j
< m
; j
++ ) a
[ 0 ] [ j
] = 1 ; for ( i
= 1 ; i
< n
; i
++ ) { for ( j
= 1 ; j
< m
; j
++ ) { if ( ( i
+ 1 ) % 2 == 0 && ( j
+ 1 ) % 2 == 0 ) a
[ i
] [ j
] = 0 ; else a
[ i
] [ j
] = a
[ i
- 1 ] [ j
] + a
[ i
] [ j
- 1 ] ; } } cout
<< a
[ n
- 1 ] [ m
- 1 ] ;
}
26. 排列字母
問題描述：
# include <iostream> # include <string> # include <algorithm> using namespace std
;
int main ( )
{ string s
; cin
>> s
; sort ( s
. begin ( ) , s
. end ( ) ) ; cout
<< s
; return 0 ;
}
27. 日期問題
題目描述：
【輸入描述】
# include <cstdio> # include <cstring> # include <iostream> # include <algorithm> # include <cmath> using namespace std
; int days
[ 13 ] = { 0 , 31 , 28 , 31 , 30 , 31 , 30 , 31 , 31 , 30 , 31 , 30 , 31 } ; bool check ( int year
, int month
, int day
)
{ if ( month
== 0 || month
> 12 ) return false ; if ( day
== 0 ) return false ; if ( month
!= 2 ) { if ( day
> days
[ month
] ) return false ; } else { int leap
= ( year
% 4 == 0 && year
% 100 != 0 ) || ( year
% 400 == 0 ) ; if ( day
> 28 + leap
) return false ; } return true ;
} int main ( )
{ int a
, b
, c
; scanf ( "%d/%d/%d" , & a
, & b
, & c
) ; for ( int i
= 19600101 ; i
<= 20591231 ; i
++ ) { int year
= i
/ 10000 ; int month
= i
/ 100 % 100 ; int day
= i
% 100 ; if ( check ( year
, month
, day
) ) { if ( year
% 100 == a
&& month
== b
&& day
== c
|| month
== a
&& day
== b
&& year
% 100 == c
|| day
== a
&& month
== b
&& year
% 100 == c
) printf ( "%d-%02d-%02d\n" , year
, month
, day
) ; } } return 0 ;
}
28. 飲料換購
題目描述：
【輸入描述】
# include <iostream> using namespace std
;
int main ( )
{ int num1
= 0 , num2
; int n
; cin
>> n
; num2
= n
; while ( n
>= 3 ) { n
-= 2 ; num1
++ ; } cout
<< num1
+ num2
; return 0 ;
}
29. 受傷的皇后
題目描述：
【輸入描述】
# include <iostream> using namespace std
;
int main ( )
{ int n
; cin
>> n
; if ( n
== 1 || n
== 2 ) { cout
<< 1 ; } else if ( n
== 3 || n
== 4 ) { cout
<< 2 ; } else if ( n
== 5 || n
== 6 ) { cout
<< 5 ; } else if ( n
== 7 || n
== 8 ) { cout
<< 7 ; } else if ( n
== 9 || n
== 10 ) { cout
<< 9 ; } return 0 ;
}
30. 最大降雨量
題目描述：
# include <iostream> using namespace std
;
int main ( )
{
cout
<< 34 << endl
;
}
31. 單詞分析
題目描述：
【輸入描述】
# include <bits/stdc++.h> using namespace std
;
int c
[ 26 ] ;
int main ( )
{ string a
; char max
; int maxcount
= 0 ; cin
>> a
; for ( int i
= 0 ; i
< a
. length ( ) ; i
++ ) { c
[ a
[ i
] - 97 ] ++ ; } for ( int i
= 0 ; i
< 26 ; i
++ ) { if ( c
[ i
] > maxcount
) { maxcount
= c
[ i
] ; max
= i
+ 97 ; } } cout
<< max
<< endl
<< maxcount
<< endl
; return 0 ;
}
32. 卡片
題目描述：
分析：
# include <iostream> using namespace std
; int main ( )
{ long int i
= 0 ; long int sum
= 2021 , tmp
= 0 ; long int b
= 0 ; for ( i
= 1 ; i
< 20210 ; i
++ ) { tmp
= i
; while ( tmp
) { b
= tmp
% 10 ; if ( b
== 1 && sum
> 0 ) { sum
-= 1 ; } tmp
/= 10 ; } if ( sum
== 0 ) { break ; } } cout
<< i
; return 0 ;
}
33. 空間
題目描述：
# include <iostream> using namespace std
;
int main ( )
{ cout
<< ( long long ) 256 * 1024 * 1024 * 8 / 32 ; return 0 ;
}
34. 時間顯示
題目描述：
【輸入描述】
# include <iostream> using namespace std
;
int main ( )
{ long long int h
; cin
>> h
; int hh
, mm
, ss
; h
= h
% ( 24 * 60 * 60 * 1000 ) ; hh
= h
/ ( 60 * 60 * 1000 ) % 24 ; mm
= h
/ ( 60 * 1000 ) % 60 ; ss
= h
/ ( 1000 ) % 60 ; printf ( "%02d:%02d:%02d\n" , hh
, mm
, ss
) ; return 0 ;
}
35. 分巧克力
題目描述：
兒童節那天有K位小朋友到小明家做客。小明拿出了珍藏的巧克力招待小朋友們。
【輸入描述】
# include <iostream> using namespace std
;
int L
[ 100005 ] ;
int W
[ 100005 ] ;
int main ( )
{ int N
, K
; cin
>> N
>> K
; for ( int i
= 1 ; i
<= N
; i
++ ) cin
>> L
[ i
] >> W
[ i
] ; int i
= 10000 ; while ( true ) { int sum
= 0 , num1
, num2
; for ( int j
= 1 ; j
<= N
; j
++ ) { num1
= L
[ j
] / i
; num2
= W
[ j
] / i
; sum
+= num1
* num2
; } if ( sum
>= K
) break ; i
-- ; } cout
<< i
; return 0 ;
}
總結
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