2369. Nim-B Sum
来源:互联网 发布:js ajax跨域的解决办法 编辑:程序博客网 时间:2024/06/05 14:53
2369. Nim-B Sum
Constraints
Time Limit: 1 secs, Memory Limit: 256 MB
Description
Note: This problem has nothing to do with siting sewage plants, power lines or wind farms. NIM is an ambigram.
The game of NIM is played with any number of piles of objects with any number of objects in each pile. At each turn, a player takes one or more (up to all) objects from one pile. In the normal form of the game, the player who takes the last object is the winner. There is a well-known strategy for this game based on the nim-2 sum.
The Nim-B sum (nim sum base B) of two non-negative integers X and Y (written NimSum(B, X, Y)) is computed as follows:
1 Write each of X and Y in base B.
2 Each digit in base B of the Nim-B sum is the sum modulo B of the corresponding digits in the base B representation of X and Y.
For example:
NimSum(2, 123, 456) = 1111011¤111001000 = 110110011 = 435
NimSum(3, 123, 456) = 11120¤121220 = 102010 = 300
NimSum(4, 123, 456) = 1323¤13020 = 10303 = 307
The strategy for normal form Nim is to compute the Nim-2 sum T of the sizes of all piles. If at any time, you end your turn with T = 0, you are guaranteed a WIN. Any opponent move must leave T not 0 and there is always a move to get T back to 0. This is done by computing NimSum(2, T, PS) for each pile; if this is less than the pile size (PS), compute the difference between the PS and the Nim-2 sum and remove it from that pile as your next move.
Write a program to compute NimSum(B, X, Y).
Input
The first line of input contains a single integer P, (1P1000), which is the number of data sets that follow. Each data set is a single line that contains the data set number, followed by a space, followed by three space separated decimal integers, B, X and Y. 2B2000000, 0X2000000, 0Y2000000.
Output
For each data set there is one line of output. It contains the data set number followed by a single space, followed by N, the decimal representation of the Nim sum in base B of X and Y.
Sample Input
4
1 2 123 456
2 3 123 456
3 4 123 456
4 5 123 456
Sample Output
1 435
2 300
3 307
4 429
Problem Source
每周一赛第一场
觉得看懂题目很重要,要记得题目要求是没有进位的,不要算成有进位的。题目中很多地方是废话,可以直接看重点。
// Problem#: 2369
// Submission#: 1984352
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <iostream>
#include <string>
#include <cmath>
using namespace std;
void NimSum(int base, int a, int b){
int al[16]={0},bl[16]={0};
int x=15;
while(a>0){
al[x]=a%base;
a/=base;
x--;
}
x=15;
while(b>0){
bl[x]=b%base;
b/=base;
x--;
}
for(x=15;x>=0;x--){
al[x]+=bl[x];
if(al[x]>=base){
al[x]%=base;
}
}
int sum=0;
for(x=15;x>=0;x--){
sum+=al[x]*pow(base,15-x);
}
cout<<sum<<endl;
}
int main(){
int n;
cin>>n;
int i;
for(i=1;i<=n;i++){
int flag,base,a,b;
cin>>flag>>base>>a>>b;
cout<<flag<<" ";
NimSum(base, a,b);
}
return 0;
}
- 2369. Nim-B Sum
- soj 2369. Nim-B Sum
- Sicily 2369. Nim-B Sum
- 31:B ·Nim-B* Sum
- Nim-sum原理及应用
- hdu1730 Northcott Game,Nim-sum
- B - Misere Nim 变式nim(LightOJ 1253)
- B. Sum of Digits
- B - Tricky Sum
- B - Equal Sum Sets
- B. Odd sum
- nim
- Nim
- Nim
- Nim
- Codeforences # 336 B B. Hamming Distance Sum
- sum a b
- codeforces 577B Modulo Sum
- HTML三种列表
- iOS开发 协议与委托(Protocol and Delegate)
- 2013腾讯编程马拉松初赛第二场(3月22日) 小Q系列故事——为什么时光不能倒流
- 关于递归超时问题的实例分析
- asp.net 文件夹如何在VS2008打开
- 2369. Nim-B Sum
- 山大哈工大C语言高校联赛
- const和volatile分析.
- 想学点真本事,不想只活在记忆里
- Ext4.0之Panel
- 2013年第二届腾讯马拉松初赛第五题
- android(一)
- Eclipse的各种快捷键
- 修复SVN错误:Revision file lacks trailing newline