LeetCode 537. Complex Number Multiplication

2018-02-06 09:42:22来源:https://www.liuchuo.net/archives/4455作者:柳婼 の blog人点击

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Given two strings representing two complex numbers.


You need to return a string representing their multiplication. Note i2 = -1 according to the definition.


Example 1:


Input: “1+1i”, “1+1i”


Output: “0+2i”


Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.


Example 2:


Input: “1+-1i”, “1+-1i”


Output: “0+-2i”


Explanation: (1 – i) * (1 – i) = 1 + i2 – 2 * i = -2i, and you need convert it to the form of 0+-2i.


Note:


The input strings will not have extra blank.


The input strings will be given in the form of a+bi, where the integer a and b will both belong to the range of [-100, 100]. And the output should be also in this form.


题目大意:给两个复数,求两个数的乘积


分析:使用sscanf和ssprinf,(m+ni)*(p+qi)=(m*p-n*q)*(n*p+m*q) i


class Solution {
public:
string complexNumberMultiply(string a, string b) {
char t[200];
int m, n, p, q;
sscanf(a.c_str(), "%d+%di", &m, &n);
sscanf(b.c_str(), "%d+%di", &p, &q);
sprintf(t, "%d+%di", (m*p-n*q), (n*p+m*q));
string ans = t;
return ans;
}
};


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