POJ 1995 Raising Module Numbers
来源:互联网 发布:三端口环形器 编辑:程序博客网 时间:2024/06/06 02:40
Raising Modulo Numbers
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 7740 Accepted: 4685
DescriptionPeople are different. Some secretly read magazines full of interesting girls’ pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market segment was so far underestimated and that there is lack of such games. This kind of game was thus included into the KOKODáKH. The rules follow:
Each player chooses two numbers Ai and Bi and writes them on a slip of paper. Others cannot see the numbers. In a given moment all players show their numbers to the others. The goal is to determine the sum of all expressions AiBi from all players including oneself and determine the remainder after division by a given number M. The winner is the one who first determines the correct result. According to the players’ experience it is possible to increase the difficulty by choosing higher numbers.
You should write a program that calculates the result and is able to find out who won the game.
Input
The input consists of Z assignments. The number of them is given by the single positive integer Z appearing on the first line of input. Then the assignements follow. Each assignement begins with line containing an integer M (1 <= M <= 45000). The sum will be divided by this number. Next line contains number of players H (1 <= H <= 45000). Next exactly H lines follow. On each line, there are exactly two numbers Ai and Bi separated by space. Both numbers cannot be equal zero at the same time.
OutputFor each assingnement there is the only one line of output. On this line, there is a number, the result of expression
(A1B1+A2B2+ … +AHBH)mod M.Sample Input
3
16
4
2 3
3 4
4 5
5 6
36123
1
2374859 3029382
17
1
3 18132
Sample Output2
13195
13
计算(A1^B1+A2^B2+ … +AH^BH)mod M
#include <iostream>#include <algorithm>#include <cstring>#include <cmath>using namespace std;typedef long long ll;ll modexp(ll a,ll b,int mod)//(a^b)%mod{ ll res = 1; while(b > 0) { if(b&1) res = res*a%mod; b >>= 1; a = a*a%mod; } return res;}int main(){ int z; cin>>z; while(z--) { int m,n; cin>>m>>n; int ans = 0; while(n--) { int ai,bi; cin>>ai>>bi; ans = (ans + modexp(ai%m,bi,m))%m; } cout<<ans<<endl; } return 0;}
- POJ 1995 Raising Module Numbers
- POJ 1995 Raising Modulo Numbers
- poj 1995 Raising Modulo Numbers
- POJ-1995-Raising Modulo Numbers
- POJ - 1995 Raising Modulo Numbers
- poj 1995 Raising Modulo Numbers
- POJ 1995 Raising Modulo Numbers
- POJ--1995--Raising Modulo Numbers
- POJ-1995 Raising Modulo Numbers
- POJ 1995 Raising Modulo Numbers
- poj 1995Raising Modulo Numbers
- POJ 1995 Raising Modulo Numbers
- POJ 1995 Raising Modulo Numbers
- POJ -1995 Raising Modulo Numbers
- Raising Modulo Numbers poj 1995
- POJ-1995 Raising Modulo Numbers
- poj 1995 Raising Modulo Numbers
- POJ 1995 Raising Modulo Numbers
- SeDebugPrivilege
- DiskLruCache理解使用心得
- opencv 图像形态学转换
- Machine Learning第二讲[多变量线性回归] -(二)计算参数分析
- ADO.NET连接字符串大全
- POJ 1995 Raising Module Numbers
- 如何用9行Python代码编写一个简易神经网络
- [More Effective C++]]指针与引用的区别
- UVa 401 Palindromes(常量数组+遍历)
- 线程范围内共享数据(二)
- Eclipse中Android项目运行时出现Unable to execute dex: java.nio.BufferOverflowException. Check the Eclipse l
- ZOJ
- OpenCV中Hough检测直线中pt1、pt2点的确定
- get,and Post请求区别