在最后一个输出后删除“+”[关闭]

Question

我有一个计算变化的 C++ 代码。

它接受输入，并根据接收变化的方式返回输出。

我需要在最后一次更改输出后删除 + 号。

有没有办法做到这一点？

我的代码：

if (n500 > 0) {
        cout << n500 << " x 500 + ";
    }
    if (n200 > 0) {
        cout << n200 << " x 200 + ";
    }
    if (n100 > 0) {
        cout << n100 << " x 100 + ";
    }
    if (n50 > 0) {
        cout << n50 << " x 50 + ";
    }
    if (n20 > 0) {
        cout << n20 << " x 20 + ";
    }
    if (n10 > 0) {
        cout << n10 << " x 10 + ";
    }
    if (n5 > 0) {
        cout << n5 << " x 5 + ";
    }
    if (n2 > 0) {
        cout << n2 << " x 2 + ";
    }
    if (n1 > 0) {
        cout << n1 << " x 1 + ";
    }
    return 0;
}

Answer 1

我会做另一种方式：当先前的显示已经完成时，放置分隔符" + "：

const char* sep = "";
if (n500 > 0) {
    std::cout << sep << n500 << " x 500";
    sep = " + ";
}
if (n200 > 0) {
    std::cout << sep << n200 << " x 200";
    sep = " + ";
}
if (n100 > 0) {
    std::cout << sep << n100 << " x 100 + ";
    sep = " + ";
}
if (n50 > 0) {
    std::cout << sep << n50 << " x 50 + ";
    sep = " + ";
}
if (n20 > 0) {
    std::cout << sep << n20 << " x 20 + ";
    sep = " + ";
}
if (n10 > 0) {
    std::cout << sep << n10 << " x 10 + ";
    sep = " + ";
}
if (n5 > 0) {
    std::cout << sep << n5 << " x 5 + ";
    sep = " + ";
}
if (n2 > 0) {
    std::cout << sep << n2 << " x 2 + ";
    sep = " + ";
}
if (n1 > 0) {
    std::cout << sep << n1 << " x 1 + ";
    sep = " + ";
}

顺便说一句，您可能会改用循环：

const int ns[9] = {
    n500, n200, n100,
    n50, n20, n10,
    n5, n2, n1
};
const char* texts[9] = {
    " x 500", " x 200", " x 100",
    " x 50", " x 20", " x 10",
    " x 5", " x 2", " x 1"
};

const char* sep = "";
for (int i = 0; i != 9; ++i) {
    if (ns[i] > 0) {
       std::cout << sep << ns[i] << texts[i];
       sep = " + ";
    }
}

Answer 2

用你的所有更新设置一个字符串，删除最后一个字符，然后用 cout 输出它。

看看字符串流?