1190 Reduced ID Numbers 求更高效的方法

最新推荐文章于 2023-10-07 17:52:59 发布

原创最新推荐文章于 2023-10-07 17:52:59 发布 · 460 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#numbers #integer #each #input #output #ini

sicily水题存档专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一个算法问题，即如何找到使学生标识号（SIN）模运算后保持唯一性的最小正整数m。通过分析输入输出样例及提供的C语言实现代码，探讨了暴力求解方法及其效率问题。

1190. Reduced ID Numbers

Total:

1418

Accepted:

581

Rating:

3.5/5.0(4 votes)


	Time Limit: 2s Memory Limit: 32MB Description T. Chur teaches various groups of students at university U. Every U-student has a unique Student Identification Number (SIN). A SIN s is an integer in the range 0 ≤ s ≤ MaxSIN with MaxSIN = 10⁶-1. T. Chur finds this range of SINs too large for identification within her groups. For each group, she wants to find the smallest positive integer m, such that within the group all SINs reduced modulo m are unique. Input On the first line of the input is a single positive integer N, telling the number of test cases (groups) to follow. Each case starts with one line containing the integer G (1 ≤ G ≤ 300): the number of students in the group. The following G lines each contain one SIN. The SINs within a group are distinct, though not necessarily sorted. Output For each test case, output one line containing the smallest modulus m, such that all SINs reduced modulo m are distinct. Sample Input Copy sample input to clipboard 2 1 124866 3 124866 111111 987651 Sample Output 1 8 我的笨方法，暴力ac #include<stdio.h> int judge(int mod[300],int n); int main() { int t,n,i,j,mod[300],mark=1; long int sin[300]; scanf("%d",&t); while(t--) { scanf("%d",&n); for(i=0;i<n;i++) scanf("%d",&sin[i]); if(n==1) {printf("%d/n",1); continue;} for(i=2;mark==1;i++) { for(j=0;j<n;j++) mod[j]=sin[j]%i; if(judge(mod,n)) mark=0; } printf("%d/n",i-1); mark=1; for(i=0;i<n;i++) mod[i]=sin[i]=0; } } int judge(int mod[300],int n) { int k,l; for(k=0;k<n;k++) for(l=k;l<n-1;l++) { if(mod[l+1]==mod[k]) return 0; } return 1; } 太慢了，用了，0.36，真不知道前面的那些牛人怎样0.01s的，求方法啊