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複習

HC01

https://ncuma-oj.math.ncu.edu.tw/problem/HC01




















// input our data
int dp[n+5] = {0};
dp[1] = 1, dp[2] = 1;
for (int i = 2 ; i < n ; i++) 
    dp[i+1] = (s[i] == s[i-2]) ? dp[i+1-2] + 1 : 1;  

for (int i = 0 ; i < q ; i++) {
    int l, r, l1, r1, l2, r2;
    cin >> l >> r;
    l1 = l + ((l & 1) == 0);
    l2 = l + (l & 1);
    r1 = r - ((r & 1) == 0);
    r2 = r - (r & 1);
    if (dp[r1] - dp[l1] + 1 == ((r1 - l1)>>1) + 1 && dp[r2] - dp[l2] + 1 == ((r2 - l2)>>1) + 1) {
        cout << "YES" << endl;
    } else {
        cout << "NO" << endl;
    }
}

函式呼叫函式 





















void IPass(){
    cout<<"PASS";
}

void IFail(){
    cout<<"FAIL";
}

void sayResult(int score){
    if (score >= 60)
        IPass();
    else
        IFail();
}

int main(){
    int score;
    cin >> score;
    sayResult(score);
}

自己呼叫自己 





void say_hello(){
    cout << "Hello~" << '\n';
    say_hello();
}

注意

必須有結束條件，或是在某些條件下才呼叫自己，這樣才有醒來的機會。

遞迴

費波那契數列

費波那契定義如下
第0項 = 0
第1項 = 1
第k項 = 第 k-1 項 + 第 k-2 項






int f(int n){
    if (n == 0) return 0;
    else if (n == 1) return 1;
    else return f(n-1) + f(n-2);
}

練習題

C003
C004

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