BC9hdoj4993&&hdoj4994 Revenge of ex-Euclid Revenge of Nim

174 阅读 0 评论 115 点赞

我是靠谱客的博主如意飞鸟，这篇文章主要介绍BC9hdoj4993&&hdoj4994 Revenge of ex-Euclid Revenge of Nim，现在分享给大家，希望可以做个参考。

Revenge of ex-Euclid

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 654 Accepted Submission(s): 380

Problem Description

In arithmetic and computer programming, the extended Euclidean algorithm is an extension to the Euclidean algorithm, which computes, besides the greatest common divisor of integers a and b, the coefficients of Bézout's identity, that is integers x and y such that ax + by = gcd(a, b).
---Wikipedia

Today, ex-Euclid takes revenge on you. You need to calculate how many distinct positive pairs of (x, y) such as ax + by = c for given a, b and c.

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case only contains three integers a, b and c.

[Technical Specification]
1. 1 <= T <= 100
2. 1 <= a, b, c <= 1 000 000

Output

For each test case, output the number of valid pairs.

Sample Input


2
1 2 3
1 1 4

Sample Output

1
3

Source

BestCoder Round #9

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=10010;
int main()
{
int t;
long long a,b,c,i,x,y;
scanf("%d",&t);
while(t--){
scanf("%lld%lld%lld",&a,&b,&c);
long long ans=0;
for(i=1;a*i<=c;++i){
if((c-a*i)%b==0&&c!=a*i)ans++;
}
printf("%lldn",ans);
}
return 0;
}

Revenge of Nim

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 522 Accepted Submission(s): 267

Problem Description

Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.
---Wikipedia

Today, Nim takes revenge on you. The rule of the game has changed a little: the player must remove the objects from the current head(first) heap. Only the current head heap is empty can the player start to remove from the new head heap. As usual, the player who takes the last object wins.

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap successively, and the player must take objects in this order, from the first to the last.

[Technical Specification]
1. 1 <= T <= 100
2. 1 <= N <= 1 000
3. 1 <= Ai <= 1 000 000 000

Output

For each test case, output “Yes” if the first player can always win, otherwise “No”.

Sample Input

Sample Output


Yes
No

Source

BestCoder Round #9

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=1010;
int num[maxn];
int main()
{
int n,m,i,j,k,t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
int cnt=0;
bool flag=false;
for(i=0;i<n;++i){
scanf("%d",&num[i]);
if(num[i]==1&&!flag){
cnt++;
}
else flag=true;
}
if(!flag){//如果全是1
if(n&1)//是奇数必胜
printf("Yesn");
else
printf("Non");
}
else {
if(cnt&1)//如果连续为1的个数为奇数则可以总是让其面对1即必须去完要么是不能再取即为输
printf("Non");
else
printf("Yesn");
}
}
return 0;
}