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  • 大数加法、乘法实现的简单版本

    有时候在外需要用大数..比如面试之类,对效率要求不高,需要的是容易记住的实现,这里有一份:

    #define big string
    #define le length()
    big addBig(big x, big y) {
        if (x.le < y.le) swap(x, y);
        // x长而y短
        for (int i = x.le - 1, j = y.le - 1; i >= 0; --i, --j) {
            if (j >= 0) x[i] += y[j] - '0';
            if (x[i] > '9') {
                x[i] -= 10;
                if (i) x[i - 1]++;
                else x = '1' + x;
            }
        }
        return x;
    }
    big mulInt(big s, int x) {
        reverse(s.begin(), s.end());
        int cmp = 0;
        for (int i = 0; i < (int)s.le; ++i) {
            cmp += (s[i] - '0') * x;
            s[i] = cmp % 10 + '0';
            cmp /= 10;
        }
        while (cmp) {
            s += cmp % 10 + '0';
            cmp /= 10;
        }
        reverse(s.begin(), s.end());
        return s;
    }
    big mulBig(big x, big y) {
        big ans;
        for (int i = y.le - 1, j = 0; i >= 0; --i, ++j) {
            big tmp = mulInt(x, y[i] - '0');
            for (int k = 0; k < j; ++k)
                tmp += '0';
            ans = addBig(ans, tmp);
        }
        return ans;
    }
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  • 原文地址:https://www.cnblogs.com/Rosebud/p/9665664.html
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