POJ - 1995 Raising Modulo Numbers
来源:互联网 发布:php银联在线支付 编辑:程序博客网 时间:2024/05/01 07:01
Description
People 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.
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.
Output
For 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
31642 33 44 55 63612312374859 30293821713 18132
Sample Output
21319513
题意:求和
思路:快速幂取模,递归和非递归两个版本
#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>typedef long long ll;using namespace std;/*int pow_mod(int a, int n, int m) { if (n == 0) return 1; if (n == 1) return a%m; int x = pow_mod(a, n/2, m); ll ans = (ll) x*x%m; if (n % 2 == 1) ans = ans * a % m; return (int)ans; } */int pow_mod(int a, int n, int m) {ll ans = 1;ll tmp = (ll)a;while (n) {if (n & 1)ans = (ans * tmp) % m;tmp = (tmp * tmp) % m;n >>= 1;}return (int) ans;}int main() {int t, n, m;while (scanf("%d", &t) != EOF && t) {for (int i = 0; i < t; i++) {int sum = 0;scanf("%d%d", &n, &m);int a, b;for (int j = 0; j < m; j++) {scanf("%d%d", &a, &b);sum = (sum + pow_mod(a, b, n)) % n;}printf("%d\n", sum);}}return 0;}
0 0
- 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
- POJ - 1995 Raising Modulo Numbers
- Oracle字符集乱码问题析及解决办法
- 批处理文件设置ip....
- iOS完整学习路线图
- 寄存器
- poj 1925 DP
- POJ - 1995 Raising Modulo Numbers
- 流VPN对于Mac用户
- 深入ARC实现机制(一)
- 你想成为优秀的Java程序员吗? 面向对象
- Costume Party poj 3663 c++
- stacked CNN深度卷积网络的简单介绍
- wireshark 包 分析 之 ftp 协议 还原 问题
- oracle 修改服务器编码
- Effective C# Item30:尽可能实现CLS兼容的程序集