first commit

2010-06-09/100.cpp

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+/*
+date:107/10/13
+author:Chen Yi Jing
+
+如果是奇數做某些事，是偶數就做其他事，有點無聊的題目
+1. i 的值是要用來跑迴圈的不能改變
+2. 題目並沒有說 start 會比 end 小 
+*/
+
+#include<iostream>
+using namespace std;
+int main() {
+    int start = 0, end = 0, tmp = 1, max = 0, value = 0;
+    int temp = 0;
+    int change = 0;
+    while(cin >> start >> end) {
+        if (start > end) {
+            temp = end;
+            end = start;
+            start = temp;
+
+            change = 1;
+        }
+            
+        for (int i = start; i <= end; i++) {
+            value = i;
+            while (value != 1) {
+                if (value % 2 != 0){
+                    value = 3 * value + 1;
+                    tmp++;
+                }
+                else {
+                    value /= 2;
+                    tmp++;
+                }
+            }
+            if(tmp > max)
+                max = tmp; 
+
+            tmp = 1; 
+        }
+
+        if (change == 1) {
+            temp = end;
+            end = start;
+            start = temp;
+        }
+        cout << start << " " << end << " " << max<< endl;
+        max = 0;
+        change = 0;
+    }
+}

2010-06-09/10012.cpp

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+/*
+date:107/10/13
+author:Chen Yi Jing
+
+要將一堆圓圈塞進一個長方形裡，求最小長方形面積
+1. 圓該怎麼擺
+*/
+
+#include <iostream>
+#include <math.h>
+using namespace std;
+int main() {
+    int line = 0, n = 0;
+
+    while(cin >> line) {
+        for (int i = 0; i < line; i++) {
+            cin >> n;
+            float num[n];
+            for (int j = 0; j < n; j++) {
+                cin >> num[j];
+            }
+            for (int j = 0; j < n; j++) {
+                cout << num[j];
+            }
+        }
+    }
+
+
+    cout<< sqrt(1.5*1.5 - 0.5*0.5);
+}

2010-06-09/10015.cpp

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+/*
+date:107/10/13
+author:Chen Yi Jing
+
+人圍成圈，用質數去繞圈，質數數完被點到的人會被殺掉，求最後倖存者
+1. TLE
+*/
+#include <iostream>
+using namespace std;
+
+int main() {
+    //將前 3500 個質數都算出來
+    int p[3502] = {0};
+    int number = 4;
+    int index = 3;
+
+    p[1] = 2;
+    p[2] = 3;
+
+    while (p[3501] == 0) {
+        for (int i = 2; i < number / 2 + 1; i++) {
+            if(number % i == 0) {
+                break;
+            }
+            if(i == number / 2) {
+                p[index] = number;
+                index++;
+            }
+        }
+        number++;
+    }
+    
+
+    //輸入的部份
+    int n = 0;
+    int copy_n = 0;
+    
+    while(cin >> n) {
+        copy_n = n;
+
+        int point = 0;  //用來看現在數到哪個人
+        int counter = 0; //用來倒數質數的
+        int num[n + 1]; //某人是否存活　１：活　０：死
+
+        //初始化為 1
+        for (int k = 0; k <= n; k++) {
+            num[k] = 1;
+        }
+
+        if (n == 0)
+            break;
+
+        while (n > 1) {   // 存活人數 > 一人
+            for (int i = 1; i < 3501; i++) {
+                if (p[i] % n !=  0) {
+                    counter = p[i] % n; //如果質數相對人數很大取餘數就好
+                }
+                else {
+                    counter = n;
+                }
+                
+                // cout << "質數　" << counter << endl;
+                while (counter > 0) {
+                    if (point == copy_n) {
+                        point = 0;
+                    }
+                    if (num[point + 1] == 1) { 
+                        counter--;
+                        point++;
+                    }
+                    else {
+                        point++;
+                    }
+
+                    if (counter == 0) {
+                        num[point] = 0;  //第 point 人被殺
+                        n = n - 1;  //總人數少 1
+                        // cout << "人數　" << n << endl;
+                    }
+                }
+                // cout << "被殺　" << point << endl;
+                // cout <<"---------------"<< endl;
+                if (n == 1)
+                    break;
+            }
+            //輸出最後倖存者
+            if (n == 1) {   
+                for (int i = 1; i <= copy_n; i++) {
+                    if (num[i] == 1) 
+                        cout << i << endl;
+                }
+            }
+            
+        }
+    }
+    cout << endl;
+}

2010-06-09/10085.cpp

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+/*
+date:107/10/13
+author:Chen Yi Jing
+
+3 * 3移動拼圖，要變成目標樣子
+*/
+#include <iostream>
+using namespace std;
+int main () {
+    
+}

2010-06-09/10579.cpp

@@ -0,0 +1,37 @@
+/*
+date:107/10/13
+author:Chen Yi Jing
+
+Fibonacci Numbers
+1. Fibonacci Numbers 要的很後面，只能另尋他路owo
+*/
+
+#include <iostream>
+using namespace std;
+int big_add () {
+    fib = n1 + n2;
+}
+int change (int (&n1)[1000]) {
+
+}
+int main () {
+    int n = 0;
+
+    while (cin >> n) {
+        int n1[1000], n2[1000], temp[1000], fib[1000];
+
+        if (n == 1)
+            fib[0] = 1;
+        else if (n == 2)
+            fib[0] = 1;
+        else 
+          while (n > 2) {
+              
+              cout << fib << endl;
+              temp = n2;
+              n2 = fib;
+              n1 = n2;
+              n--;
+          }  
+    }
+}

2010-06-09/10664.cpp

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+/*
+date:107/10/13
+author:Chen Yi Jing
+
+給很多數字，要判斷這些數字是否能分成兩堆等大的數字
+1.沒有說一行幾個數字：StringStream 做字串的分割（空白切割）
+2.動態規劃 0 1 背包的問題：可以用0-1背包的思想来进行解决。
+先将n个数全部累加起来，总数为sum，这里和0-1背包的不同之处在于，这里的价值和重量是一样的，
+所以将一个包的容量设置为sum/2,然后你就采用0-1背包思路从n个物品中尽可能的选择数字，
+使得数字的和最接近sum/2,那么这个解就是最优解。
+*/
+#include <iostream>
+#include <sstream>
+
+using namespace std;
+int main() {
+    int n = 0;
+    cin >> n;
+
+    for (int i = 0; i < n; i++) {
+        int temp = 0, index = 0;
+        int weight[20] = {0};
+        int sum = 0, str = 0;
+        string input;
+
+
+        getline(cin, input);    //取得一行的輸入
+        stringstream ss (input);  //StringStream 做字串的分割（空白切割）
+        while(ss >> str) {
+            weight[index] = str;
+            index++;
+        }
+        
+        for (int j = 0; j < index; j++) {
+            sum += weight[j];
+        }
+        if (sum % 2 != 0) {
+            cout << "NO" << endl;
+        }
+        else {
+            cout << "YES" << endl;
+        }
+        cout << sum << endl;
+
+        //清空 ss
+        ss.str("");
+        ss.clear();
+    }
+}
+