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#include<cstdio>
#include<iostream>
using namespace std;
long long gcd(long long a, long long b){
if(b == 0)return a;
else return gcd(b, a%b);
}
int main(){
int T;
cin >> T;
while(T--){
long long g,l;
int flag = 0;
cin >> l >> g;
if(g%l!=0){cout << -1 << endl; continue;}
if(l == 1){cout << l << " " << g << endl; continue;}
long long a,b;
for(long long i = 1; i*l <=g ; i++){
if(g%(i*l)!=0)continue;
a = i*l;
b = g/i;
if(gcd(g,i*l) != l)continue;
flag = 1;
break;
}
if(flag)cout << min(a,b) << " " << max(a,b) <<endl;
else cout << -1 << endl;
}
return 0;
}
//自己测数据的时候才发现原来最小公倍数一定是最大公因数的倍数，也就是说只要判断最小公倍数是否为最大公因数的倍数即可

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#include<cstdio>
#include<iostream>
using namespace std;
long long gcd(long long a, long long b){
if(b == 0)return a;
else return gcd(b, a%b);
}
int main(){
int T;
cin >> T;
while(T--){
long long A,C;
cin >> A >> C;
if(C%A!=0){cout << "NO SOLUTION" << endl;continue;}
long long B = C/A;
long long g = gcd(B, A);
while(g>1){
B*=g;
A/=g;
g = gcd(A,B);
}
cout << B << endl;
}
return 0;
}
//正如我们所知道的A*B= gcd(A,B)*lcm(A,B);
//所以B=gcd(A,B)*(lcm(A,B)/A);

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#include<cstdio>
#include<iostream>
#include<cstring>
#define mod 1000000
using namespace std;
int dp[210][210];
int main(){
long long n,k;
dp[0][0] = 1;
for(int i = 1; i <= 200; i++){
dp[i][0] = 1;
for(int j = 1; j<= i ; j++){
dp[i][j] = dp[i-1][j-1]+dp[i-1][j];
dp[i][j]%=mod;
}
}
while(cin >> n >> k && n*k){
cout << dp[n+k-1][k-1] << endl;
}
return 0;
}
//记忆化了C(n,k),哟哟哟；

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#include<cstdio>
#include<iostream>
#include<cstring>
#define INF 100000000
using namespace std;
int main(){
int T,t=0;
cin >> T;
while(t++<T){
int n,m;
cin >> m >> n;
int i = 2,ans = INF;
while(m!=1){
int p = 0;//质因子个数;
while(m%i==0){
m/=i;
p++;
}
if(p){
int num = n, temp = 0;
while(num){
temp+=num/i; // 每除一次有i的个数
num/=i;
}
ans = min(ans, temp/p);//取最小的才能满足所有的,temp/p自己意会比如p=3,i=2;
}
i++;
}
printf("Case %d:n",t);
if(ans) cout << ans << endl;
else cout << "Impossible to divide" << endl;
}
return 0;
}
//质因数分解

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//暴力代码
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int m;
long long biao[100000];
void init(int T){
m = 0;
for(long long i = 1; i*i<=T; i++){
if(T%i == 0){
biao[m++] = i;
if(T/i!=i)
biao[m++] = T/i;
}
}
sort(biao, biao+m);
}
long long gcd(long long a, long long b){
return b==0?a : gcd(b,a%b);
}
long long solve(long long T){
int ans = 0;
for(int i = 0; i < m; i++){
for(int j = i; j < m; j++){
if(biao[i]*biao[j]/gcd(biao[i],biao[j]) == T)
ans++;
}
}
return ans;
}
int main(){
int T;
while(cin >> T && T){
init(T);
cout << T << " " << solve(T) << endl;
}
return 0;
}
//数论代码
#include<cstdio>
#include<iostream>
using namespace std;
int main(){
long long n,nn;
while(cin >> n && n){
nn = n;
long long ans = 1;//未有质因数时
for(long long i = 2; i*i <= nn; i+= 2){
int cot = 0;
while(nn%i==0){
cot++;
nn/=i;
}
ans*=(cot*2+1);
if(i == 2)i--;
}
if(nn>1) ans = ans *3; //是否有剩下一个比sqrt(T)更大的数
ans = (ans+1)/2;
cout << n << " " << ans << endl;
}
return 0;
}

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 #include<cstdio>
#include<iostream>
#include<cstring>
#include<set>
using namespace std;
unsigned long long maxn = ~0LL >> 1;
int biao1[] = {2 ,3 ,5 ,7 ,11 ,13 ,17 ,19 ,23 ,29 ,31 ,37 ,41 ,43 ,47 ,53 ,59 ,61};
int biao2[66];
set<unsigned long long> a;
int main(){
for(int i = 0;biao1[i]; i++)biao2[biao1[i]]=1;
for(unsigned long long i = 2;; i++){
unsigned long long cnt= - 1,x = maxn;
while(x){
x/=i;
cnt++;
}
if(cnt<4)break;
unsigned long long temp = i*i;
for(int j = 2;j <= cnt; j++ ){
if(!biao2[j])a.insert(temp);
temp*= i ;
}
}
a.insert(1);
for(set<unsigned long long>::iterator it = a.begin()++; it != a.end(); ++it){
cout << *it << endl;
}
return 0;
}

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#include<cstdio>
#include<iostream>
#include<cstring>
#define CLR(a,v) memset(a,v, sizeof a)
using namespace std;
int a[10];
void init(){
CLR(a,0);
}
long long f[20]={1,1,2};
long long factorial(int a){
if(f[a])return f[a];
else return f[a] = factorial(a-1)*a;
}
int main(){
int n;
while(cin >> n && n){
init();
int temp;
for(int i = 0;i < n; i++){
cin >> temp;
a[temp]++;
}
unsigned long long ans = 0;
for(int i = 0; i <= 9 ; i++){
if(a[i] <= 0)continue;
long long t=factorial(n-1);
a[i]--;
for(int j = 0; j <= 9 ; j++){
t /= factorial(a[j]);
}
a[i]++;
ans+=t*i;
}
unsigned long long an = ans;
for(int i = 1; i < n; i++)
ans = ans*10+an;
cout << ans << endl;
}
return 0;
}

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#include<cstdio>
#include<iostream>
#include<cstring>
#define MOD 1000000007
using namespace std;
long long quickpow(long long b, long long n){
long long ans=1;
while(n){
if(n&1)ans = ans * b%MOD;
b = b*b%MOD;
n>>=1;
}
return ans;
}
int main(){
int n=0,T;
cin >> T;
while(n++<T){
long long b;
cin >> b;
printf("Case #%d: %lldn",n,b*quickpow(2,b-1)%MOD);
}
return 0;
}

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#include<cstdio>
#include<iostream>
#include<cstring>
#define CLR(a,v) memset(a,v, sizeof a)
using namespace std;
int a[3];
char ans[2] = {'S','T'};
int main(){
int T,n=0;
cin >> T;
getchar();
while(n++<T){
CLR(a,0);
char ch;
while((ch = getchar())!= 'n' ){
a[(ch-'0')%3]++;
}
//cout << a[0] <<a[1] << a[2] <<endl;
if((a[1]+a[2]*2)%3!=0 && !(max(a[1],a[2])%3==2 && min(a[1],a[2])==0))a[0]++;//拿走一个使他能被三整除
a[0]%=2;
printf("Case %d: %cn",n,ans[(a[0]+1)%2]);
}
return 0;
}

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#include<cstdio>
#include<iostream>
#include<cstring>
#define CLR(a,v) memset(a,v,sizeof a)
using namespace std;
int biao[39]={0};
void init(){
biao[1] = biao[2] = 9;
for(int i = 3; i<= 38; i++){
biao[i] = biao[i-2]*10;
}
}
int ans[40];
void fuck(int n){
int k = n, len = 0;
while(k>biao[len+1]){
k-=biao[++len];
}
len++;
int temp=1;
for(int i = 1; i < (len+1)/2 ; i++)temp*=10;
temp += k-1;
int count = 0;
while(temp){
int a = temp %10;
ans[len/2 + count] = ans[(len-1)/2-count] = a;
temp/=10;
count++;
}
for(int i = 0; i < len; i++)
cout << ans[i];
cout << endl;
}
int main(){
init();
int n;
while(cin >> n && n)fuck(n);
return 0;
}
#include<cstdio>
#include<iostream>
#include<cstring>
#define CLR(a,v) memset(a,v,sizeof a)
using namespace std;
int biao[39]={0};
void init(){
biao[1] = biao[2] = 9;
for(int i = 3; i<= 38; i++){
biao[i] = biao[i-2]*10;
}
}
int ans[40];
void fuck(int n){
int k = n, len = 0;
while(k>biao[len+1]){
k-=biao[++len];
}
len++;
int temp=1;
for(int i = 1; i < (len+1)/2 ; i++)temp*=10;
temp += k-1;
int count = 0;
while(temp){
int a = temp %10;
ans[len/2 + count] = ans[(len-1)/2-count] = a;
temp/=10;
count++;
}
for(int i = 0; i < len; i++)
cout << ans[i];
cout << endl;
}
int main(){
init();
int n;
while(cin >> n && n)fuck(n);
return 0;
}

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#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int main(){
long long T,t=0;
while(cin >> T && T){
long long ans = 0;
int flag = 0 ;
for(long long i = 2 ; i <= sqrt(T)+1; i++){
if(T%i!=0)continue;
flag ++;
int temp = 1;
while(T%i == 0){
temp*=i;
T/=i;
}
ans+=temp;
}
if(flag==0)ans+=T+1;
else if(flag==1)ans+=T;
else if(T>1)ans+=T;
printf("Case %lld: %lldn",++t,ans);
}
return 0;
}

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#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int biao[400000]={0,0,1,1};
int main(){
int n;
cin >> n;
while(n--){
int t;
cin >> t;
int m = (t-1)>>1;
int h = m*(m+1);
if(t&1) h -= m;
cout << h << endl;
}
}