Question

似乎无法让这个工作。这个想法是计算每次计数后头尾的百分比，在每次迭代后累积。除了我一直在 nan％进行计算。有谁看到我做错了什么？

void flipCoin(time_t seconds, int flipCount){
    vector<int> flips;
    float headCount = 0;
    float tailCount = 0;
    double headProbability = double((headCount/(headCount + tailCount))*100);
    double tailProbability = double((tailCount/(headCount + tailCount))*100);

    for (int i=0; i < flipCount; i++) {
        int flip = rand() % (HEADS - TAILS + 1) + TAILS;
        flips.push_back(flip);
        if (flips[i] == 1) {
            tailCount++;
            cout << "Tail Percent: " << tailProbability << "%" << endl;
        }else{
            headCount++;
            cout << "Head Percent: " << headProbability << "%" << endl;
        }
    }
}

Answer 1

你想这样做

void flipCoin(time_t seconds, int flipCount){
    vector<int> flips;
    float headCount = 0;
    float tailCount = 0;
    double headProbability = 0;
    double tailProbability = 0;

    for (int i=0; i < flipCount; i++) {
        int flip = rand() % (HEADS - TAILS + 1) + TAILS;
        flips.push_back(flip);
        if (flips[i] == 1) {
            tailCount++;
            tailProbability = double((tailCount/(headCount + tailCount))*100);
            cout << "Tail Percent: " << tailProbability << "%" << endl;
        }else{
            headCount++;
            headProbability = double((headCount/(headCount + tailCount))*100);
            cout << "Head Percent: " << headProbability << "%" << endl;
        }
    }
}

此外，如果您只想打印百分比，则不需要tailProbability或headProbablity。您可以直接打印计算。

编辑：你也可以用i + 1替换headCount + tailCount。但如果稍后更改功能，则可能会出现问题。

Answer 2

您的问题在于您只计算headProbability和tailProbability一次 - 当您计算它们时，它将评估为0 / 0。

float headCount = 0;
float tailCount = 0;
// Both are 0, end up as 0/0
double headProbability = double((headCount/(headCount + tailCount))*100);  
double tailProbability = double((tailCount/(headCount + tailCount))*100);

//Never recalculated in the loop

要解决此问题，只需将它们最初定义为0，然后在循环中重新计算它们。你似乎也不需要vector，因为你只需要跟踪计数，而不是每次翻转的结果：

double headProbability = 0;
double tailProbability = 0;

for(int i = 0; i < flipCount; ++i) {
    int flip = rand() % (HEAS - TAIL + 1) + TAILS; //This could be simplified
    if(flip == 1) {
        ++tailCount;
        //Recalculate tailProbability
    } else { 
       //Etc...
    }
}

Answer 3

nan表示不是数字。你得到它是因为你除以0。

你的代码计算概率只有一次，在任何翻转硬币之前。您需要在循环中放置calculs，以便在每次迭代后重新计算概率。

void flipCoin(time_t seconds, int flipCount){
    vector<int> flips;
    float headCount = 0;
    float tailCount = 0;
    double headProbability;
    double tailProbability;

    for (int i=0; i < flipCount; i++) {
        int flip = rand() % (HEADS - TAILS + 1) + TAILS;
        flips.push_back(flip);

        headProbability = double((headCount/(headCount + tailCount))*100);
        tailProbability = double((tailCount/(headCount + tailCount))*100);

        if (flips[i] == 1) {
            tailCount++;
            cout << "Tail Percent: " << tailProbability << "%" << endl;
        }else{
            headCount++;
            cout << "Head Percent: " << headProbability << "%" << endl;
        }
    }
}

C ++：计算每次迭代期间的概率百分比

3 个答案: