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  1. #include<stdio.h>
  2. #include<stdlib.h>
  3.  
  4. int main()
  5. {
  6.     int x,y,t,a,b,ans,flag,i,j;
  7.     scanf("%d",&t);
  8.     // READ THE COMMENT IN THE LOOP AND COME BACK HERE
  9.     // I HAVE JUST TAKEN THIS FROM MY CODE
  10.     //YOU CAN DO IT MANUALLY
  11.     //IT IS BETTER THAT YOU DO IT YOURSELF
  12.     long long int arr[10000]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003};
  13.      int sum=0;
  14.     while(t--)
  15.     {
  16. 		sum=0;
  17.         scanf("%d %d",&x,&y);
  18.         // I AM MAKING SOME DIFFERENCES IN YOUR CODE. I AM SURE YOU WILL UNDERSTAND IT :)
  19.         // THE MAIN LOGIC OF THE QUESTION IS THAT WE HAVE TO FIND THE FIRST PRIME NUMBER AFTER THE SUM (X+Y) AND PRINT THE DIFF BETWEEN (FIRST PRIME AFTER X+Y)-(X+Y)
  20.         // WE CAN DO IT IN 2 WAYS-> WE CAN FIND THE NEXT PRIME IN TEST CASE THAT IS AFTER ACCEPTING X AND Y.
  21.         // SECOND WAY IS -> WE CAN PRECOMPUTE THE PRIMES NUMBERS AND STORE IN AN ARRAY. THIS WILL BE MORE EFFICIENT AS THE RUNNING TIME IS CONCERNED.
  22.         // I EXPECT THAT YOU KNOW HOW TO FIND WHETHER A NUMBER IS PRIME OR NOT?
  23.         // IF THEN PUT A LOOP FROM 1-5000 AND STORE ALL THE NUMBERS WHICH ARE PRIME IN AN ARRAY.
  24.         // I AM TAKING THE LIMIT AS 5000 BECAUSE THE CONSTRAINTS GIVEN IN THE PROBELEM IS "X AND Y"<=1000. IF THEN THE TOTAL SUM WILL NOT EXCEED 2000. SO WE NEED TO CALCULATE ONLY PRIMES UPTO THE NEXT PRIME NUMBER AFTER 2000
  25.         // YOU HAVE TO PRE-COMPUTE THE PRIMES OUTSIDE THE  WHILE LOOP-ELSE THAT WILL BE A WASTE. YOU WILL BE PRE-COMPUTING EACH TIME.
  26.         // SUPPOSE YOU KNOW HOW TO FIND ALL THE PRIME NUMBERS BEWTEEN THE ABOVE RANGE AND MOVING TO THE NEXT STEP
  27.         //MAKE A LINEAR SEARCH IN THE ARRAY SUCH THAT FIND THE FIRST NUMBER THAT IS GREATER THAN X+Y.
  28.         //BREAK THE LOOP AND PRINT THE DIFFERENCE BETWEEN THAT NUMBER AND X+Y. YOUR ANSWER WILL BE ACCEPTED :)
  29.         int sum=x+y;
  30.          for(int i=0;i<sizeof(arr)/arr[0];i++)
  31.             {
  32.               if(arr[i]>sum)
  33.               {
  34.                 ans=arr[i]-sum;
  35.                 break;
  36.               } 
  37.  
  38.             }
  39.           printf("%d\n",ans);
  40.  
  41.     }
  42.     return 0;
  43. }
  44.  
  45.  
Success #stdin #stdout 0s 3352KB
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https://ideone.com/bYNcih
language:
C++14 (gcc 8.3)
created:
9 years ago
visibility:
secret

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