2012 ACM/ICPC Asia Regional Tianjin Online

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1455 Accepted Submission(s): 440

Problem Description

　　Haoren is very good at solving mathematic problems. Today he is working a problem like this:
　　Find three positive integers X, Y and Z (X < Y, Z > 1) that holds
　　 X^Z + Y^Z + XYZ = K
　　where K is another given integer.
　　Here the operator “^” means power, e.g., 2^3 = 2 * 2 * 2.
　　Finding a solution is quite easy to Haoren. Now he wants to challenge more: What’s the total number of different solutions?
　　Surprisingly, he is unable to solve this one. It seems that it’s really a very hard mathematic problem.
　　Now, it’s your turn.

Input

　　There are multiple test cases.
　　For each case, there is only one integer K (0 < K < 2^31) in a line.
　　K = 0 implies the end of input.
　　

Output

　　Output the total number of solutions in a line for each test case.

Sample Input

95360

Sample Output

110　　
Hint
9 = 1^2 + 2^2 + 1 * 2 * 253 = 2^3 + 3^3 + 2 * 3 * 3
 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

貌似昨天rp都用光了，今天上来A题秒过n多人，压力山大，然后看到上面那倒，一看就是二分查找的感觉，但是输入2^31，电脑就卡了不给输出，开始还以为难道超时了，难道还有什么高深莫测的优化，无奈放下，刚看到最后一题简单想做来着，不知道我们那个账号有的人这么多尽然a了，简单的都被A完了我干嘛Y-Y......然后看到这道题过的人越来越多，果断再来debug，一个一个数据输到104800，还有输出，到104900，瞬间卡屏，不至于瞬间超时，然后终于想通了，数据溢出了（Y-Y),虽然k《=2^31；二分计算函数值的时候会超过的啊Y-Y,终于闷骚的打了一瓶酱油过了，昨天过一道，今天还是一道Y——Y，下次来个突破吧，阿弥陀佛善哉善哉

#include<string.h>#include<stdio.h>#include<math.h>long long  y,z,k,sum,f,ok[1290];long long fx(long  long a,long long b,long long c){ long long i,aa=1,bb=1; for (i=1;i<=c;i++) {    aa=aa*a;//本来用pow，记得好像是计算实数会不会有精度误差不用算了-    bb=bb*b; } return aa+bb+a*b*c;}void find(long long  l,long long r){ long long m=(l+r)/2,key; key=fx(m,y,z); if (key==k) {if (m>0&&m<y) f=1; return ;} if (l>=r) return; if (key<k) find(m+1,r);       else find(l,m-1);}int main(){ long long t,exp=0; while (scanf("%lld",&k),k) {  sum=0;  t=sqrt(k);  if (t*t==k)//z=2，直接开方，  {   if (t%2==1) sum=t/2;       else sum=t/2-1;  } for (z=3;fx(1,2,z)-exp<=k;z++) {  y=2;  while (fx(1,y,z)-exp<=k)//说起exp，是为了调试程序用的看边界条件，其实边界开始就对的最后exp=0，可有可无  {   f=0;   find(1,y-1);   sum=sum+f;   ++y;  } } printf("%lld\n",sum); } return 0;}最后再去hdu用gcc跑了下15ms，再用c跑了下0ms，冲到第一了吼吼。