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#include<iostream> using namespace std; long diedai(int n) { long result; long p_result; long n_result; result=p_result=1; //这一段表达的斐波拉契数列第n项的值 while(n>2) { n-=1; n_result=p_result;//把前一项的值赋给前一项的前一项 p_result=result; // result=p_result+n_result;//结果等于前一项加上前一项的前一项 } return result; } int main() { for (int i = 1; i < 10; i++) cout << diedai(i) << endl; }
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