USACO Arithmetic Progressions 【构造等差数列】-白红宇

USACO Arithmetic Progressions 【构造等差数列】

阅读量：5235 次

发布时间：2019-06-14

本文共 1966 字，大约阅读时间需要 6 分钟。

USER: Jeremy Wu [wushuai2]TASK: ariprogLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.005 secs, 11880 KB]   Test 2: TEST OK [0.008 secs, 11880 KB]   Test 3: TEST OK [0.008 secs, 11876 KB]   Test 4: TEST OK [0.014 secs, 11880 KB]   Test 5: TEST OK [0.016 secs, 11880 KB]   Test 6: TEST OK [0.065 secs, 11880 KB]   Test 7: TEST OK [0.378 secs, 11880 KB]   Test 8: TEST OK [0.818 secs, 11880 KB]   Test 9: TEST OK [0.802 secs, 11876 KB]All tests OK.

Your program ('ariprog') produced all correct answers! This is your submission #5 for this problem. Congratulations!

还是算蛮简单的一道构造题目= = 差点以为是要去搜索了...

/*ID: wushuai2PROG: ariprogLANG: C++*///#pragma comment(linker, "/STACK:16777216") //for c++ Compiler#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #define Max(a,b) (((a) > (b)) ? (a) : (b))#define Min(a,b) (((a) < (b)) ? (a) : (b))#define Abs(x) (((x) > 0) ? (x) : (-(x)))#define MOD 1000000007#define pi acos(-1.0)using namespace std;typedef long long ll ;typedef unsigned long long ull ;typedef unsigned int uint ;typedef unsigned char uchar ;template
                  
                    inline void checkmin(T &a,T b){ if(a>b) a=b;}template
                   
                     inline void checkmax(T &a,T b){ if(a
                    
                     > n >> m; init(); sort(a, a + len); for(i = 0; i < len; ++i){ int num = a[i]; for(j = 0; j < len; ++j){ int cur = a[j]; //fisrt a + b int b = cur - num; if(b * (n - 1) + num > a[len - 1]) break; if(b < 1) continue; for(k = 2; k < n; ++k){ if(!status[num + k * b]){ break; } } if(n == k){ ans[llen].a = num; ans[llen++].b = b; } } } if(!llen){ fout << "NONE" << endl; return 0; } sort(ans, ans + llen, cmp); for(i = 0; i < llen; ++i){ cout << ans[i].a << ' ' << ans[i].b << endl; fout << ans[i].a << ' ' << ans[i].b << endl; } fin.close(); fout.close(); return 0;}