int M[15];
unsigned long long nn,m;
unsigned long long gcd(unsigned long long a,unsigned long long b ){
return b == 0?a:gcd(b,a%b);
}
unsigned long long lcm(unsigned long long a,unsigned long long b){
return (a*b)/(gcd(a,b));
}
unsigned long long ...n1 = d1*q1+r1
n2 = d1Xq2+r1
n3 = d1Xq3+r1
要找出d1出來
a1= n1-n2 = d1X(q1-q2)
a2 = n1-n3 = d1X(q1-q3)
只要找出a1跟a2的最大公因數即可求出
如果只有兩個呢？
a1 = (n1-n2)X1 = d1X(q1-q2)
那麼n1-n2自然就是最大的那一個答案了。
int gcd(int a,int b){
return b == 0?a:gcd(b,a%b);
}
int main(int argc, char const* argv[])
{
string ...#define happy 1
#define unhappy 2
int main(int argc, char const* argv[])
{
//1000000000
//999999999
//81*9 = 729
int n[730] = {0} ;
n[1] = happy;
for(int i = 2;i<= 729;i++){
if(n[i]){
continue;
}
set<int> tmp;
tmp.insert(i);
int sum = 0;
int g = i;
bool ...int N[50005];
void binarySearch(int search,int n){
int left = 0,right = n-1;
while(left <= right){
int mid = (left+right)/2;
if(N[mid]<search){
left = mid+1;
}else if(N[mid]>search){
right = mid-1;
}else if(N[mid] == search){
if(mid+1>n-1 ...int gcd(int a,int b){
return b == 0?a:gcd(b,a%b);
}
int main(int argc, char const* argv[])
{
bool prime[65540];
long long ppp[65540];
int power[65540];
int m = 0;
for(long long i = 2;i<65540;i++){
if(!prime[i]){
ppp[m++] = i ...int main(int argc, char const* argv[])
{
int t;
while(scanf("%d",&t) != EOF){
while(t--){
unsigned long long x,k;
scanf("%llu %llu",&x,&k);
int a = floor((double)x/k);
int b = ceil((double)x/k);
unsigned long long p;
unsigned long long q;
bool ans = false ...int f91(int N){
if(N >= 101){
return N-10;
}
if(N <= 100){
return f91(f91(N+11));
}
}
int main(int argc, char const* argv[])
{
int n;
while(scanf("%d",&n) && n != 0){
printf("f91(%d) = %d\n",n,f91(n));
}
return 0;
}
int main(int argc, char const* argv[])
{
int n;
while(~scanf("%d",&n) && n != 0){
int nn = sqrt(n);
int count = 0;
int k = n;
for(int i = 2;i<= nn;i++){
if(n%i == 0){
while(n%i == 0) {
n = n/i;
}
count++;
}
}
if(n>1){
count++;
}
n ...int main(int argc, char const* argv[])
{
bool prime[2001];
prime[0] = true;
prime[1] = true;
for(int i = 2;i<2001;i++){
if(!prime[i]){
for(int j = i*i;j<2001;j = j+i){
prime[j] =true;
}
}
}
int p;
char str[2001];
while(scanf("%d",&p) != EOF ...#define MAX 500001
int N[MAX];
int solution(int n){
if(N[n])
return N[n];
if(n % 2 == 0){
N[n] = solution(n/2)*2;
}
else{
N[n] = (solution((n+1)/2)-1)*2;
}
return N[n];
}
int main(int argc, char const* argv[])
{
int n;
N[1 ...