bool isprime[32769];
int prime[32769];
typedef unsigned int ui;
int main(int argc, char const* argv[])
{
int m = 0;
for(int i = 2;i <= 32768;i++ ){
if(!isprime[i]){
prime[m++] = i;
for(int j = i*i;j <= 32768;j = j+i)
isprime[j] = true;
}
}
string line;
while ...int n,m;
unsigned long long combination(){
unsigned long long p = 1;
if(m>(n/2)){
m = n-m;
}
for(int i = 1;i <= m;i = i+1){
p = p*(n-i+1)/i;
}
return p;
}
int main(int argc, char const* argv[])
{
while(~scanf("%d %d",&n,&m ...int row[105];
int col[105];
int main(int argc, char const* argv[])
{
string line;
int n;
while(~scanf("%d",&n) && n != 0){
getchar();
int tmp = n;
int rowe= 0;
int cole = 0;
int rr = 0;
int ic,ir;
while(tmp--){
getline(cin,line);
istringstream stream(line);
int bit ...bool prime[65541];
long long u[33300];
int main(int argc, char const* argv[])
{
prime[1] = true;
int m = 0;
for(long long int i = 2;i <= 65540;i++){
if(!prime[i]){
u[m++] = i;
for(long long int j = i*i;j <= 65540;j = j+i){
prime[j ...int main(int argc, char const* argv[])
{
int n;
int i = 0;
int r[50];
while(~scanf("%d",&n) && n != 0){
int sum = 0;
int p = n;
int j = 0;
while(n--){
int h;
scanf("%d",&h);
sum = sum+h;
r[j++] = h;
}
int avg = sum/p;
int ans ...典型的0/1背包問題。
int main(int argc, char const* argv[])
{
int n;
int m;
int dp[2000];
int put[25][2000];
while(scanf("%d",&n) != EOF){
vector<int> cd;
memset(dp,0,sizeof(dp));
memset(put,0,sizeof(put));
scanf("%d",&m);
while(m--){
int l;
scanf("%d",&l);
cd.push_back ...這題用BFS即可解出，有一點必須要注意的是，雖然題目要求當經過的routers數量相同，則輸出最小的id，但因為題目的input已經有幫我們做好排序，所以每次找到的自然都會是最小的id。
//ignore header files
bool bfs(int*,int,int,vector<int>*);
void print_path(int *,int ,int);
int main(int argc, char const* argv[])
{
int n;
char line[1005];
while(scanf("%d",&n) != EOF){
getchar();
vector <int> list[305];
for(int i = 0;i<n;i ...int main(int argc, char const* argv[])
{
int q[8] = {9,8,7,6,5,4,3,2};
int t;
while(~scanf("%d",&t)){
while(t--){
long long n;
scanf("%lld",&n);
int sum = 0;
int k = 1;
int test = 1;
int tmp = n;
if(n != 1){
for(int ...在 Worker Thread Pattern 中，worker thread 會依序抓一件工作來執行。當沒有工作時，worker thread 會停下來等待新的工作過來。
worker thread 也有人稱為 Background Thread。也有人把視點放在管理 worker thread 的地方，稱之為 thread pool。
程式範例：
Client 會建立 Requset，傳給 Channel。例如 ClientThread 類別。
Channel 從 Client 取得 Request，傳給 Worker。例如 Channel 類別。
Worker 會從 Channel ...
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