POJ 1207 求最大数链长度 暴力枚举数学题

这个题目直接用的暴力枚举，但是还是WA了几次

原因是这句话You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j. 

注意i可能大于j，此时需要交换顺序，但是输出的时候还会是原始顺序

比如 输入1 10 输出 1 10 20 ；输入10 1 输出 10 1 20

因此交换i和j需要用标志位记录一下

Source Code

Problem: 1207		User: yangliuACMer
Memory: 244K		Time: 16MS
Language: C++		Result: Accepted

//暴力枚举 //21:22 #include <iostream> using namespace std; int main(){ int i,j,k,count,max,newk,temp; bool flag; while(cin>>i>>j){ max = 1; flag = true; if(i > j){ temp = i; i = j; j = temp; flag = false; } for(k = i; k <= j; k++){ count = 1; newk = k;//注意计算循环节的时候k的值会发生变化 while(newk != 1){ count++; if(newk % 2) newk = 3 * newk + 1; else newk/=2; } if(count > max) max = count; } if(flag) cout<<i<<" "<<j<<" "<<max<<endl; else cout<<j<<" "<<i<<" "<<max<<endl; } return 0; } //21:42