solution.md

# 夺冠概率
足球比赛具有一定程度的偶然性，弱队也有战胜强队的可能。

假设有甲、乙、丙、丁四个球队。根据他们过去比赛的成绩，得出每个队与另一个队对阵时取胜的概率表:
```
    甲   乙  丙  丁   
甲   -  0.1 0.3 0.5
乙 0.9  -   0.7 0.4 
丙 0.7  0.3 -   0.2
丁 0.5  0.6 0.8 -
```
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。

![](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。

 

## 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

```

## 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;
}
```