HDU 3579 Hello Kiki 解题报告(中国剩余定理 非互质)
来源:互联网 发布:linux怎么安装pycharm 编辑:程序博客网 时间:2024/05/22 06:04
Hello Kiki
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1835 Accepted Submission(s): 649
Problem Description
One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing "门前大桥下游过一群鸭,快来快来 数一数,二四六七八". And then the cashier put the counted coins back morosely and count again...
Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
One day Kiki's father found her note and he wanted to know how much coins Kiki was counting.
Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
One day Kiki's father found her note and he wanted to know how much coins Kiki was counting.
Input
The first line is T indicating the number of test cases.
Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi
Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi
Output
For each case output the least positive integer X which Kiki was counting in the sample output format. If there is no solution then output -1.
Sample Input
2214 575 56519 54 40 24 8011 2 36 20 76
Sample Output
Case 1: 341Case 2: 5996
Author
digiter (Special Thanks echo)
解题报告:在刘汝佳老师的书上学了中国剩余定理,不过限制是给定的m[]都是两两互质的。本题中的除数就不是互质的,所以我们要换一种方法:合并模方程。
关于合并的证明,我在这里看到的:http://yzmduncan.iteye.com/blog/1323599/。建议大家自己证明一下。
另外,题目有个坑爹的地方,输出正数。如果结果是0,就加上最后的模数。
本人代码如下:
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100;int mi[maxn];int ai[maxn];int cas=1;void gcd(int a,int b,int &d,int &x,int &y){ if(b==0) { d=a; x=1; y=0; } else { gcd(b,a%b,d,y,x); y-=x*(a/b); }}int china_2(int n,int m[],int a[]){ int aa = a[0]; int mm = m[0]; for(int i=1;i<n;i++) { int ta=((a[i]-aa)%m[i]+m[i])%m[i]; int d,x,y; gcd(mm,m[i],d,x,y); if(ta%d) return -1; int newM = mm/d*m[i]; aa = ((long long)ta/d*x*mm%newM+aa)%newM; mm = newM; } aa = (aa+mm)%mm; if(aa==0) aa=mm; return aa;}void work(){ int n; scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d", mi+i); for(int i=0;i<n;i++) scanf("%d", ai+i); int ans = china_2(n,mi,ai); printf("Case %d: %d\n", cas++, ans);}int main(){ int T; scanf("%d",&T); while(T--) work();}
0 0
- HDU 3579 Hello Kiki 解题报告(中国剩余定理 非互质)
- HDU 3579 Hello Kiki(中国剩余定理)
- HDU 3579 Hello Kiki(拓展中国剩余定理)
- HDU 3579 Hello Kiki 中国剩余定理(合并方程
- HDU 3579 Hello Kiki 中国剩余定理 不互质情况
- Hdu 1573 X问题 + Hdu 3579 Hello Kiki (模线性方程组-非互质中国剩余定理)
- HDU 3579 Hello Kiki【中国剩余】
- hdu 3579 Hello Kiki 中国剩余定理(不互质形式)模板题
- Hello Kiki hdu 3579 求解一元线性同余方程组ps:中国剩余定理
- hdoj-3579-Hello Kiki【中国剩余定理 & 除数不互质】
- Hello Kiki(hdu3579+不互质的中国剩余定理)
- HDOJ 题目3579 Hello Kiki(中国剩余定理,拓展的欧几里得)
- HDU 5446 Unknown Treasure 解题报告(Lucas定理 + 中国剩余定理)
- hdu 3579(中国剩余定理非互质)
- 【中国剩余定理-M不互质的情况】HDU Hello Kiki 2579
- 中国剩余定理(非互质)hdu3579 Hello Kiki ====注意刚好全部整除的情况
- POJ 1006 Biorhythms 解题报告(中国剩余定理)
- hdu 3579(中国剩余定理)
- Windows下后缀为*.zip.001文件的解压方法[转]
- URAL 1353. Milliard Vasya's Function(dp)
- 如何查看window文件系统块大小
- 在gitcafe下用hexo建的新博客
- JAVA核心技术 第五章 继承 类、超类、子类 受保护的对象
- HDU 3579 Hello Kiki 解题报告(中国剩余定理 非互质)
- 如何在XAMPP中设置多个网站
- 设置jdk环境变量时lib中的rt.jar ,dt.jar ,tool.jar是什么 ,作用是什么
- SOS 使用总结
- 2010水仙花
- URAL 1119. Metro(简单dp)
- 方法和函数的区别?
- ORA-12514 TNS 监听程序当前无法识别连接描述符中请求服务 的解决方法
- Python模块学习 ---- zlib 数据压缩流(内存数据)