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先不说题,我要吐槽!!!这题太坑了,你给了1-1,000,000的范围,然后你竟然不用任何精妙的算法,直接暴力循环就能AC,然而我还以为要用什么精妙的算法,写了大半天。。。(虽然最后还没写出来,手动黑脸),原因是忘了比较i和j的大小顺序,最后气哄哄的跑去看题解,f**k,竟然真的是水题,我还是太菜狗了。
正题:
没有废话,直接上代码:
//problems://Problems in Computer Science are often classified as belonging //to a certain class of problems (e.g., NP, Unsolvable, Recursive).// In this problem you will be analyzing a property of an algorithm //whose classification is not known for all possible inputs. //Consider the following algorithm: // 1. input n // 2. print n // 3. if n = 1 then STOP // 4. if n is odd then n <- 3n + 1 // 5. else n <- n / 2 // 6. GOTO 2 //Given the input 22, the following sequence of numbers will be printed //22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 ////It is conjectured that the algorithm above will terminate //(when a 1 is printed) for any integral input value. Despite the //simplicity of the algorithm, it is unknown whether this conjecture is true. //It has been verified, however, for all integers n such that 0 < n < 1,000,000 //(and, in fact, for many more numbers than this.) ////Given an input n, it is possible to determine the number of numbers printed //(including the 1). For a given n this is called the cycle-length of n. //In the example above, the cycle length of 22 is 16. //For any two numbers i and j you are to determine the maximum cycle //length over all numbers between i and j. //Input://The input will consist of a series of pairs of integers i and j, //one pair of integers per line. All integers will be less than 1,000,000 and greater than 0. ////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. ////You can assume that no opperation overflows a 32-bit integer. //Output://For each pair of input integers i and j you should output i, j, //and the maximum cycle length for integers between and including i and j. //These three numbers should be separated by at least one space with all //three numbers on one line and with one line of output for each line of input. //The integers i and j must appear in the output in the same order in which //they appeared in the input and should be followed by the maximum cycle length (on the same line). #include总结:我是菜鸡!!!using namespace std;//求n得循环长度 int find(int n){ int cnt=0; while(n!=1){ cnt++; if(n%2) n=3*n+1; else n/=2; } return cnt+1;}int main(){ int i,j,max; while(cin>>i>>j){ //你没有看错,我的那个拙略的算法,就是将i,j都放进一个数组中,然后遍历数组求出最大长度 int vis[1000001]={0}; max=0; cout< <<' '< <<' '; //趁i,j位置没变,先偷偷输出他 //将i,j按大小排列 if(i>j){ int t=i; i=j; j=t; } //遍历i到j for(int k=i;k<=j;k++){ if(!vis[k]){ //若k的长度未知 int cnt=find(k); vis[k]=cnt; if(k%2 && k*3+1<=j) vis[k*3+1]=cnt-1;//当k长度确定时,且k为奇数时,3k+1的长度确定 else if(k*2<=j) vis[2*k]=cnt+1; //当k确定时,2k的长度确定 } if(vis[k]>max) max=vis[k]; //求最大长度 } cout< <