1.题目介绍
2.题解
至于为何\(C_{3}^{0}+C_{3}^{1}+C_{3}^{2}+C_{3}^{3} = 2^{3}\)
可以使用数学归纳法：
1.对于\(C_{0}^{0} = 2^0\)
2.假设对于n = k，\(C_{k}^{0}+C_{k}^{1}+C_{k}^{2}+...+C_{k}^{k} = 2^{k}\)成立
3.对于n = k + 1，
\(C_{k+1}^{0}+C_{k+1}^{1}+C_{k+1}^{2}+...+C_{k+1}^{k}+C_{k+1}^{k+1} = C_{k+1}^0+(C_k^0+C_k^1)+(C_k^1+C_k^2)+...+(C_k^{k-1}+C_k^k)+C_{k+1}^{k+1}\)
代码
提供了三种输出方式
#include <iostream>
#include <cmath>
#include <iomanip>
int main() {
char s;
int number, n = 0;
long long result=0;
while (std::cin >> number) {
result += number;
n++;
}
std::cout << std::fixed << std::setprecision(0) << result * pow(2, n-1) << std::endl;
//std::cout << (long long) result * pow(2, n - 1) << std::endl; //存在使用e输出法的问题
printf("%lld\n", (long long)(result * pow(2, n-1) ));
result *= pow(2, n - 1);
printf("%lld", result);
return 0;
}