csv to token "1,2,3" => [1,2,3]

std::string input = "abc,def,ghi";
std::istringstream ss(input);
std::string token;

while(std::getline(ss, token, ',')) {
    std::cout << token << '\n';
}

source:

http://stackoverflow.com/questions/11719538/how-to-use-stringstream-to-separate-comma-separated-strings

std::strtok



http://en.cppreference.com/w/cpp/string/byte/strtokk

跑者的修練

#include <iostream>

using namespace std;

int main()

{

      int T;

      cin>>T;

      long long M,N;

      while(T--)

      {

           cin>>M>>N;

           long long ans=(1+N)*N/2;

           if(N>M)

                 ans+=(((N-M)+(N%M))*(int)(N/M))/2;

           cout<<ans<<endl;

      }

      return 0;

}

b024: 指南宮的階梯 (luke)

1. /**********************************************************************************/  
2. /*  Problem: b024 "指南宮的階梯" from 動態規劃-爬樓梯問題                                        */  
3. /*  Language: C++                                                                 */  
4. /*  Result: AC (4ms, 184KB) on ZeroJudge                                          */  
5. /*  Author: luke at 2014-08-23 15:38:35                                           */  
6. /**********************************************************************************/  
7.   
8. #include <iostream>  
9. #include <iomanip>  
10. #include <stdio.h>  
11. using namespace std;  
12. long long int a[100];  
13. long long int solve( int number){  
14.     if(a[number]!=0){  
15.         return a[number];  
16.     }  
17.     if(number==1){  
18.         a[1]=1;  
19.         return 1;  
20.     }  
21.     if(number==2){  
22.         a[2]=2;  
23.         return 2;  
24.     }  
25.     else{a[number]=solve(number-1)+solve(number-2);  
26.         return solve(number-1)+solve(number-2);}  
27. }  
28. int main () {  
29.     int m;  
30.     cin>>m;  
31.     cout<<solve(m)<<" ";  
32.     long long int k=solve(m);  
33.     int v=k%m;  
34.     cout<<solve(v);  
35.       
36.       
37.        return 0;  
38. }  

b021 "指考分發" (skipper)

1. /**********************************************************************************/  
2. /*  Problem: b021 "指考分發" from 排序-複合欄位                                             */  
3. /*  Language: C++                                                                 */  
4. /*  Result: AC (4ms, 200KB) on ZeroJudge                                          */  
5. /*  Author: skipper at 2014-08-22 10:30:20                                        */  
6. /**********************************************************************************/  
7.   
8. #include <iostream>  
9. #include <algorithm>  
10. #include <numeric>  
11. using namespace std;  
12. struct Score {  
13.     int no,subj[5],total;  
14. };  
15. bool compare (Score a,Score b) {   
16.     if (a.total>b.total) return true;      
17.     if (a.total==b.total) return a.subj[2]>=b.subj[2];  
18.     return false;   
19. }  
20.   
21. int main ()  
22. {  
23.     int N; cin >> N;  
24.     Score *s;  
25.     s= new Score[N];  
26.     for (int i=0;i<N;i++){  
27.         cin >> s[i].no; s[i].total = 0;  
28.         for (int j=0;j<5;j++) { cin >> s[i].subj[j];}  
29.         s[i].total=accumulate(s[i].subj,s[i].subj+5,0);  
30.     }  
31.       
32.     sort(s,s+N,compare);  
33.     for (int i=0;i<N;i++){  
34.         cout << s[i].no << endl;  
35.     }  
36. }  

執行時間測量(gcd)

#include <time.h>

#include <iostream>

#include <math.h>

#include <algorithm>

using namespace std;

int gcd1(int a, int b){

     for (int d = 1; d <= min(a, b); d++){

         if (a%d == 0 && b%d == 0) return d;

     }

}

int gcd2(int a, int b){

     if (a<b) swap(a, b);

     if (a%b)

         return b;

     else

         return gcd2(a - b, b);

}

// source: http://www.math.wustl.edu/~victor/mfmm/compaa/gcd.c

/* Standard C Function: Greatest Common Divisor */

int gcd(int a, int b)

{

     int c;

     while (a != 0) {

         c = a; a = b%a;  b = c;

     }

     return b;

}

/* Recursive Standard C Function: Greatest Common Divisor */

int gcdr(int a, int b)

{

     if (a == 0) return b;

     return gcdr(b%a, a);

}

void main(){

     clock_t begin = clock();

     // your CODE

     srand(clock());

     int x = rand();

     int y = rand();

     for (int i = 0; i < 100000000; i++){

         gcd1(x, y);

     }

     clock_t end = clock();  

     cout << end - begin;

     //system("pause");

}

NPSC2008國中組決賽 C.喵喵捉老鼠 (Queue 的應用)

g047

#include <iostream>
#include <string>
#include <map>
#include <queue>
using namespace std;
int grid[100][100],N,M;
map<char,int> tr;
enum {WALL=-3,MOUSE,SPACE,CAT};
struct Position {int r; int c;};
Position k; 

queue<Position> q;
Position moves[4]={{0,1},{1,0},{0,-1},{-1,0}};
bool FOUND;
int steps;
int main(){
 tr['#']=WALL, tr['@']=MOUSE, tr['.']=SPACE, tr['K']=CAT;
 while(cin>>N && N>0){
  for (int i=0;i<N;i++){
   string row; cin >> row;
   M = row.length();
   for (int j=0;j<M;j++){
    grid[i][j]=tr[row[j]];
    if (row[j]=='K') {k.r=i; k.c=j;};
   }
  }
  FOUND = false;
  while(!q.empty()) q.pop();
  q.push(k);
  while(!q.empty()&&(!FOUND)){
   k=q.front();q.pop();
   steps = grid[k.r][k.c]+1;
   for (int i=0;i<4 &&(!FOUND);i++){
    Position p = {k.r+moves[i].r, k.c+moves[i].c};
    switch(grid[p.r][p.c]){
    case MOUSE:
     FOUND=true;     
     break;
    case SPACE:
     grid[p.r][p.c]=steps;
     q.push(p);
    }
   }
  }
  if(FOUND)
   cout << steps << endl;
  else
   cout << "= =\"" << endl;
 }
}