update exercises

data/1.算法初阶/1.蓝桥杯/堆的计数/solution.cpp

@@ -44,7 +44,7 @@ int main()
     {
         if (i * 2 + 1 <= n)
         {
-            dp[i] = (C(s[i] - 1, s[i * 2 + 1]) * dp[i * 2 + 1]) % mod * dp[i * 2] % mod;
+            dp[i] = (C(s[i], s[i * 2]) * dp[i * 2]) % mod * dp[i * 2] % mod;
         }
     }
     cout << dp[1] << endl;

data/1.算法初阶/1.蓝桥杯/堆的计数/solution.md

@@ -56,7 +56,31 @@ CPU消耗  < 1000ms
 ## aop
 ### before
 ```cpp
-
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+const int maxn = 100000;
+ll s[maxn + 10], dp[maxn + 10], inv[maxn + 10], f[maxn + 10];
+ll n;
+ll mod = 1000000009;
+ll qPow(ll a, ll b)
+{
+    ll ans = 1;
+    while (b != 0)
+    {
+        if (b & 1)
+        {
+            ans = ans * a % mod;
+        }
+        a = a * a % mod;
+        b >>= 1;
+    }
+    return ans % mod;
+}
+ll C(ll n, ll m)
+{
+    return (f[n] * inv[f[m]]) % mod * inv[f[n - m]] % mod;
+}
 ```
 ### after
 ```cpp
@@ -65,21 +89,125 @@ CPU消耗  < 1000ms
 
 ## 答案
 ```cpp
-
+int main()
+{
+    cin >> n;
+    f[0] = 1;
+    for (int i = 1; i <= maxn; i++)
+    {
+        f[i] = (f[i - 1] * i) % mod;
+        inv[i] = qPow(i, mod - 2);
+    }
+    for (int i = n; i >= 1; i--)
+    {
+        s[i] = (s[i * 2 + 1] <= n ? s[i * 2 + 1] : 0) + (s[i * 2] <= n ? s[i * 2] : 0) + 1;
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        dp[i] = 1;
+    }
+    for (int i = n; i >= 1; i--)
+    {
+        if (i * 2 + 1 <= n)
+        {
+            dp[i] = (C(s[i] - 1, s[i * 2 + 1]) * dp[i * 2 + 1]) % mod * dp[i * 2] % mod;
+        }
+    }
+    cout << dp[1] << endl;
+    return 0;
+}
 ```
 ## 选项
 
 ### A
 ```cpp
-
+int main()
+{
+    cin >> n;
+    f[0] = 1;
+    for (int i = 1; i <= maxn; i++)
+    {
+        f[i] = (f[i - 1] * i) % mod;
+        inv[i] = qPow(i, mod - 2);
+    }
+    for (int i = n; i >= 1; i--)
+    {
+        s[i] = (s[i * 2 + 1] <= n ? s[i * 2 + 1] : 0) + (s[i * 2] <= n ? s[i * 2] : 0) + 1;
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        dp[i] = 1;
+    }
+    for (int i = n; i >= 1; i--)
+    {
+        if (i * 2 + 1 <= n)
+        {
+            dp[i] = (C(s[i] - 1, s[i * 2 - 1]) * dp[i * 2 + 1]) % mod * dp[i * 2] % mod;
+        }
+    }
+    cout << dp[1] << endl;
+    return 0;
+}
 ```
 
 ### B
 ```cpp
-
+int main()
+{
+    cin >> n;
+    f[0] = 1;
+    for (int i = 1; i <= maxn; i++)
+    {
+        f[i] = (f[i - 1] * i) % mod;
+        inv[i] = qPow(i, mod - 2);
+    }
+    for (int i = n; i >= 1; i--)
+    {
+        s[i] = (s[i * 2 + 1] <= n ? s[i * 2 + 1] : 0) + (s[i * 2] <= n ? s[i * 2] : 0) + 1;
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        dp[i] = 1;
+    }
+    for (int i = n; i >= 1; i--)
+    {
+        if (i * 2 + 1 <= n)
+        {
+            dp[i] = (C(s[i] - 1, s[i * 2 - 1]) * dp[i * 2 - 1]) % mod * dp[i * 2] % mod;
+        }
+    }
+    cout << dp[1] << endl;
+    return 0;
+}
 ```
 
 ### C
 ```cpp
-
+int main()
+{
+    cin >> n;
+    f[0] = 1;
+    for (int i = 1; i <= maxn; i++)
+    {
+        f[i] = (f[i - 1] * i) % mod;
+        inv[i] = qPow(i, mod - 2);
+    }
+    for (int i = n; i >= 1; i--)
+    {
+        s[i] = (s[i * 2 + 1] <= n ? s[i * 2 + 1] : 0) + (s[i * 2] <= n ? s[i * 2] : 0) + 1;
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        dp[i] = 1;
+    }
+    for (int i = n; i >= 1; i--)
+    {
+        if (i * 2 + 1 <= n)
+        {
+            dp[i] = (C(s[i], s[i * 2]) * dp[i * 2]) % mod * dp[i * 2] % mod;
+        }
+    }
+    cout << dp[1] << endl;
+    return 0;
+}
 ```

data/1.算法初阶/1.蓝桥杯/大数乘法/solution.md

 # 大数乘法
 对于32位字长的机器，大约超过20亿，用int类型就无法表示了，我们可以选择int64类型，但无论怎样扩展，固定的整数类型总是有表达的极限！如果对超级大整数进行精确运算呢？一个简单的办法是：仅仅使用现有类型，但是把大整数的运算化解为若干小整数的运算，即所谓：“分块法”。
 
-![](https:
+![](https://img-blog.csdn.net/20160125091111485)  
 上图表示了分块乘法的原理。可以把大数分成多段（此处为2段）小数，然后用小数的多次运算组合表示一个大数。可以根据int的承载能力规定小块的大小，比如要把int分成2段，则小块可取10000为上限值。注意，小块在进行纵向累加后，需要进行进位校正。
 
 

data/1.算法初阶/1.蓝桥杯/夺冠概率/desc.md

@@ -10,7 +10,9 @@
 ```
 数据含义：甲对乙的取胜概率为0.1，丙对乙的胜率为0.3，...
 
-现在要举行一次锦标赛。双方抽签，分两个组比，获胜的两个队再争夺冠军。（参见【1.jpg】）
+现在要举行一次锦标赛。双方抽签，分两个组比，获胜的两个队再争夺冠军。
+
+![](https://img-blog.csdn.net/20150228234310457?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvd3IxMzI=/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center)
 
 请你进行10万次模拟，计算出甲队夺冠的概率。
 

data/1.算法初阶/1.蓝桥杯/夺冠概率/solution.cpp

-#include <stdio.h>
-#include <stdlib.h>
-int main(void)
-{
-    int i, j, k; //count
-    int randfigure;
-    int awin = 0;
+#include <iostream>
+using namespace std;
+double rate[4][4] = {
+    {0, 0.1, 0.3, 0.5},
+    {0.9, 0, 0.7, 0.4},
+    {0.7, 0.3, 0, 0.2},
+    {0.5, 0.6, 0.8, 0}};
 
-    for (i = 0; i < 100000; i++)
+int main()
+{
+    int i, j, k;
+    double time, sum;
+    time = 0;
+    sum = 0;
+    for (i = 1; i < 4; i++)
     {
-        randfigure = rand();
-        if (randfigure % 3 == 0)
-        { //AB together
-            if (randfigure % 10 < 1)
-            { //A win B
-                if (randfigure % 10 < 2)
-                { //C win D
-                    if (randfigure % 10 < 3)
-                    { //A win C
-                        awin++;
-                    }
-                }
-                else
-                { //D win C
-                    if (randfigure % 10 < 5)
-                    { //A win D
-                        awin++;
-                    }
-                }
-            }
-        }
-        else if (randfigure % 3 == 1)
-        { //AC together
-            if (randfigure % 10 < 3)
-            { //A win C
-                if (randfigure % 10 < 4)
-                { //B win D
-                    if (randfigure % 10 < 1)
-                    { //A win B
-                        awin++;
-                    }
-                }
-                else
-                { //D win B
-                    if (randfigure % 10 < 5)
-                    { //A win D
-                        awin++;
-                    }
-                }
-            }
-        }
-        else
-        { //AD together
-            if (randfigure % 10 < 5)
-            { //A win D
-                if (randfigure % 10 < 3)
-                { //C win B
-                    if (randfigure % 10 < 3)
-                    { //A win C
-                        awin++;
-                    }
-                }
-                else
-                { //B win C
-                    if (randfigure % 10 < 1)
-                    { //A win B
-                        awin++;
-                    }
+        for (j = 1; j < 4; j++)
+        {
+            for (k = 1; k < 4; k++)
+            {
+                if (j != i && k != i && j != k)
+                {
+                    sum += rate[0][i] * rate[j][k];
                 }
             }
         }
+        time++;
     }
-    printf("%f\n", awin / 100000.0);
+    cout << sum / time;
+    return 0;
 }
\ No newline at end of file

data/1.算法初阶/1.蓝桥杯/夺冠概率/solution.md

@@ -11,7 +11,9 @@
 ```
 数据含义：甲对乙的取胜概率为0.1，丙对乙的胜率为0.3，...
 
-现在要举行一次锦标赛。双方抽签，分两个组比，获胜的两个队再争夺冠军。（参见【1.jpg】）
+现在要举行一次锦标赛。双方抽签，分两个组比，获胜的两个队再争夺冠军。
+
+![](https://img-blog.csdn.net/20150228234310457?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvd3IxMzI=/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center)
 
 请你进行10万次模拟，计算出甲队夺冠的概率。
 
@@ -20,7 +22,13 @@
 ## aop
 ### before
 ```cpp
-
+#include <iostream>
+using namespace std;
+double rate[4][4] = {
+    {0, 0.1, 0.3, 0.5},
+    {0.9, 0, 0.7, 0.4},
+    {0.7, 0.3, 0, 0.2},
+    {0.5, 0.6, 0.8, 0}};
 ```
 ### after
 ```cpp
@@ -29,21 +37,109 @@
 
 ## 答案
 ```cpp
-
+int main()
+{
+    int i, j, k;
+    double time, sum;
+    time = 0;
+    sum = 0;
+    for (i = 1; i < 4; i++)
+    {
+        for (j = 1; j < 4; j++)
+        {
+            for (k = 1; k < 4; k++)
+            {
+                if (j != i && k != i && j != k)
+                {
+                    sum += rate[0][i] * rate[j][k] * rate[0][j];
+                }
+            }
+        }
+        time++;
+    }
+    cout << sum / time;
+    return 0;
+}
 ```
 ## 选项
 
 ### A
 ```cpp
-
+int main()
+{
+    int i, j, k;
+    double time, sum;
+    time = 0;
+    sum = 0;
+    for (i = 1; i < 4; i++)
+    {
+        for (j = 1; j < 4; j++)
+        {
+            for (k = 1; k < 4; k++)
+            {
+                if (j != i && k != i && j != k)
+                {
+                    sum += rate[0][i] * rate[0][j];
+                }
+            }
+        }
+        time++;
+    }
+    cout << sum / time;
+    return 0;
+}
 ```
 
 ### B
 ```cpp
-
+int main()
+{
+    int i, j, k;
+    double time, sum;
+    time = 0;
+    sum = 0;
+    for (i = 1; i < 4; i++)
+    {
+        for (j = 1; j < 4; j++)
+        {
+            for (k = 1; k < 4; k++)
+            {
+                if (j != i && k != i && j != k)
+                {
+                    sum += rate[j][k] * rate[0][j];
+                }
+            }
+        }
+        time++;
+    }
+    cout << sum / time;
+    return 0;
+}
 ```
 
 ### C
 ```cpp
-
+int main()
+{
+    int i, j, k;
+    double time, sum;
+    time = 0;
+    sum = 0;
+    for (i = 1; i < 4; i++)
+    {
+        for (j = 1; j < 4; j++)
+        {
+            for (k = 1; k < 4; k++)
+            {
+                if (j != i && k != i && j != k)
+                {
+                    sum += rate[0][i] * rate[j][k];
+                }
+            }
+        }
+        time++;
+    }
+    cout << sum / time;
+    return 0;
+}
 ```

data/1.算法初阶/1.蓝桥杯/等腰三角形/solution.md

@@ -4,7 +4,7 @@
 先用1,2,3，…的自然数拼一个足够长的串
 用这个串填充三角形的三条边。从上方顶点开始，逆时针填充。
 比如，当三角形高度是8时：  
-![](https:
+![](https://img-blog.csdnimg.cn/20190203174943179.png)
 
 输入，一个正整数n(3<n<300),表示三角形的高度  
 输出，用数字填充的等腰三角形。  
@@ -15,20 +15,20 @@
 5
 
 #### 程序应该输出：
-![](https:
+![](https://img-blog.csdnimg.cn/20190203175048400.png)
 
 ### 再例如：
 #### 输入：
 10
 
 #### 程序应该输出：
-![](https:
+![](https://img-blog.csdnimg.cn/20190203175133907.png)
 ### 再例如：
 #### 输入：
 15
 
 #### 程序应该输出：
-![](https:
+![](https://img-blog.csdnimg.cn/20190203175213126.png)
 
 
 ## aop

data/1.算法初阶/1.蓝桥杯/颠倒的价牌/solution.md

@@ -26,15 +26,15 @@ using namespace std;
 ```cpp
 int main()
 {
-    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
-    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
-    int profit1[1111][2];                //利润1
-    int profit2[1111][2];                //利润2
-    int before_reverse;                  //颠倒前
-    int after_reverse;                   //颠倒后
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; 
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; 
+    int profit1[1111][2];                
+    int profit2[1111][2];                
+    int before_reverse;                  
+    int after_reverse;                   
     int i = 0;
     int j = 0;
-    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
+    for (int a = 1; a < 7; a++) 
     {
         for (int b = 0; b < 7; b++)
         {
@@ -44,16 +44,16 @@ int main()
                 {
                     before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
                     after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
-                    //因为是颠倒之后所以第一位可以是0
+                    
                     if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
-                    //在200-300范围内，我们以正数为准
+                    
                     {
                         profit1[i][0] = before_reverse;
                         profit1[i][1] = after_reverse - before_reverse;
                         i++;
                     }
                     else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
-                    //在800-900范围内
+                    
                     {
                         profit1[j][0] = before_reverse;
                         profit2[j][1] = after_reverse - before_reverse;
@@ -84,15 +84,15 @@ int main()
 ```cpp
 int main()
 {
-    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
-    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
-    int profit1[1111][2];                //利润1
-    int profit2[1111][2];                //利润2
-    int before_reverse;                  //颠倒前
-    int after_reverse;                   //颠倒后
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; 
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; 
+    int profit1[1111][2];                
+    int profit2[1111][2];                
+    int before_reverse;                  
+    int after_reverse;                   
     int i = 0;
     int j = 0;
-    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
+    for (int a = 1; a < 7; a++) 
     {
         for (int b = 0; b < 7; b++)
         {
@@ -102,16 +102,16 @@ int main()
                 {
                     before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
                     after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
-                    //因为是颠倒之后所以第一位可以是0
+                    
                     if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
-                    //在200-300范围内，我们以正数为准
+                    
                     {
                         profit1[i][0] = before_reverse;
                         profit1[i][1] = after_reverse - before_reverse;
                         i++;
                     }
                     else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
-                    //在800-900范围内
+                    
                     {
                         profit1[j][0] = before_reverse;
                         profit2[j][1] = after_reverse - before_reverse;
@@ -141,15 +141,15 @@ int main()
 ```cpp
 int main()
 {
-    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
-    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
-    int profit1[1111][2];                //利润1
-    int profit2[1111][2];                //利润2
-    int before_reverse;                  //颠倒前
-    int after_reverse;                   //颠倒后
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; 
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; 
+    int profit1[1111][2];                
+    int profit2[1111][2];                
+    int before_reverse;                  
+    int after_reverse;                   
     int i = 0;
     int j = 0;
-    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
+    for (int a = 1; a < 7; a++) 
     {
         for (int b = 0; b < 7; b++)
         {
@@ -159,16 +159,16 @@ int main()
                 {
                     before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
                     after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
-                    //因为是颠倒之后所以第一位可以是0
+                    
                     if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
-                    //在200-300范围内，我们以正数为准
+                    
                     {
                         profit1[i][0] = before_reverse;
                         profit1[i][1] = after_reverse - before_reverse;
                         i++;
                     }
                     else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
-                    //在800-900范围内
+                    
                     {
                         profit1[j][0] = before_reverse;
                         profit2[j][1] = after_reverse - before_reverse;
@@ -198,15 +198,15 @@ int main()
 ```cpp
 int main()
 {
-    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
-    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
-    int profit1[1111][2];                //利润1
-    int profit2[1111][2];                //利润2
-    int before_reverse;                  //颠倒前
-    int after_reverse;                   //颠倒后
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; 
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; 
+    int profit1[1111][2];                
+    int profit2[1111][2];                
+    int before_reverse;                  
+    int after_reverse;                   
     int i = 0;
     int j = 0;
-    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
+    for (int a = 1; a < 7; a++) 
     {
         for (int b = 0; b < 7; b++)
         {
@@ -216,16 +216,16 @@ int main()
                 {
                     before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
                     after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
-                    //因为是颠倒之后所以第一位可以是0
+                    
                     if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
-                    //在200-300范围内，我们以正数为准
+                    
                     {
                         profit1[i][0] = before_reverse;
                         profit1[i][1] = after_reverse - before_reverse;
                         i++;
                     }
                     else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
-                    //在800-900范围内
+                    
                     {
                         profit1[j][0] = before_reverse;
                         profit2[j][1] = after_reverse - before_reverse;