HDU1017 A Mathematical Curiosity(模拟)

注意输入问题和每一块之间需要输出空行

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <string>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '\n'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long int ll;
typedef __int64 bi;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){
    
    return x&(-x);}
ll gcd(ll a,ll b){
    
    return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){
    
    if(!b){
    
    d=a,x=1,y=0;}else{
    
    ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){
    
    ll res=1;a%=MOD;while(b>0){
    
    if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){
    
    return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){
    
    ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){
    
     ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {
    
    if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = {
    
     {
    
    1,0}, {
    
    -1,0},{
    
    0,1},{
    
    0,-1} };


signed main(void)
{
    
    
	int t;
	cin>>t;
	int n,m;
	while(t--){
    
    
		int c = 0;
		while(cin>>n>>m){
    
    
			if(n == 0 && m == 0) break;
			int cnt = 0;
			for(int a = 1;a<n;++a)
				for(int b = a+1;b<n;++b)
					if((a*a+b*b+m)%(a*b) == 0)
						++cnt;
			printf("Case %d: %d\n", ++c,cnt);
		}
		if(t!=0) cout<<endl; 
	}
}

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转载自blog.csdn.net/YSJ367635984/article/details/113958255