C++:计算每次迭代期间的概率百分比

Question

似乎无法让它工作。这个想法是计算每次计数后正面和反面的百分比，在每次迭代后累积。除了我的计算不断得到 nan%。有人看到我做错了什么吗？

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结果。

编辑:您也可以将 headCount + tailCount 替换为 i+1 。但如果您稍后更改功能，这可能会导致问题。