for( 起始條件 ; 判斷式 ; 條件運算 ){ // 迴圈內容 } while( 判斷式 ){ // 迴圈內容 } do{ // 迴圈內容 } while( 判斷式 ) ;

Example: Please output 0 ~ 9 for( int i = 0; i <10 ; i++){ cout << i << “ ”; } int i = 0; while( i < 10){ cout << i << “ ”; i++; } 請按任意鍵繼續... _

while & do while int i = 10; while( i < 10){ cout << i << “ ”; i++; } int i = 10; do{ cout << i << “ ”; i++; } while( i < 10); 請按任意鍵繼續... _ 10 請按任意鍵繼續... _

起始條件、判斷式、條件運算可以不只一個。 for(int i = 0, j = 1, k = 0; j < 5 || i++ ; j++, k++){ cout<< i << " " << j << " " << k << endl; } 請按任意鍵繼續..._

What about for(int i = 0, j = 1, k = 0; j < 5 || ++i ; j++, k++){ cout<< i << " " << j << " " << k << endl; } Infinite loop

WHY ?

while( true ) while( false ) = while( 1) = while( 0) 若條件為 true 可以省略條件不寫 → while() 若條件為 false 可以省略迴圈不寫 → 留他何用？

隨手一個 break 救救無限迴圈

for(int i = 0 ; ; i++) { if(i == 10){ break; } cout << i << " "; } 請按任意鍵繼續... _ break 會跳出一層迴圈

for(int i = 0; i < 10; i++) { if(i % 3 == 0){ continue; } cout << i << " "; } continue 會直接跳到最後面 請按任意鍵繼續... _

(condition1 || condition2 && condition3) condition1 is true condition1 is false → condition2 is true → condition3 is true→ true → condition2 is true→ condition3 is false→ false → condition2 is false→ false → true

NOTE int array[3] = {0, 1, 2}; for( int i = 0; i <= 3; i++) { cout << array[i] << " "; } 請按任意鍵繼續... _

NOTE for( int i = 0; i <= 3; i++) { cout << i << " "; } cout << i << " "; name lookup of `i' changed for new ISO `for' scoping

NOTE Give appropriate condition. Infinite loop would cause Time Limit Exceeded. for( int i = 0; i < 10; i--) { cout << i << " "; }

NOTE while( condition 1){ … while( condition 2){ … while( condition 3){ … } … } … } 括號的對應很重要。

Recursion

遞迴 → 自己呼叫自己 Example: n! int f( int n){ if( n == 1){ return 1; } else return n * f( n-1); } n! = n * ( n - 1) * … * 1 = n * ( n - 1)!

int f( 3){ if( 3 == 1){ return 1; } else return 3 * f( 3-1); } When n = 3 ： int f( 2){ if( 2 == 1){ return 1; } else return 2 * f( 2-1); } int f( 1){ if( 1 == 1){ return 1; } else return 1 * f( 1- 1); }

int f( 3){ if( 3 == 1){ return 1; } else return 3 * f( 3-1); } When n = 3 ： int f( 2){ if( 2 == 1){ return 1; } else return 2 * 1; }

int f( 3){ if( 3 == 1){ return 1; } else return 3 * 2; } When n = 3 ： 注意： 一定要寫終止條件

費氏數列 … F(1) = 1F(2) = 1F( n) = F( n - 1 ) + F( n - 2) int f( int n ){ if ( n == 1 || n == 2){ return 1; } else{ return f( n - 1 ) + f( n - 2 ); }

F(5) F(1) = 1F(2) = 1F( n) = F( n - 1 ) + F( n - 2) F(4)F(3) F(2) F(1) F(2) F(1)

int GCD( int x, int y){ if( x % y != 0){ return GCD( y, x % y); } else{ return y; } 最大公因數 GCD GCD( 187, 209) GCD( 209, 187) GCD( 187, 22) GCD( 22, 11) int GCD( 187, 209){ if( 187 % 209 != 0){ return GCD( 209, 187); } else{ return y; } int GCD( 209, 187){ if(209 % 187 != 0){ return GCD( 187, 22); } else{ return y; } int GCD( 187, 22){ if( 187 % 22 != 0){ return GCD( 22, 11); } else{ return y; } int GCD( 22, 11){ if( 22 % 11 != 0){ return GCD( y, x % y); } else{ return 11; } 11

int f( n){ if( n == 1){ return 1; } else return n * f( n-1); } for( int i = 1, j = 1; i <= n; i++){ j = j * i; } Recursion Loop

Exercise

基礎 : 100 The 3n + 1 problem 408 Uniform Generator 488 Triangle Wave F the 9s 進階 : Queens Chess Problem 785 Grid Colouring Sweet Child Makes Trouble