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Solution of BLOPER spoj - spoj Operators

Maximum & Minimum Sum possible with N natural number is n*(n+1)/2 and out of it we need to take S. then we'll left with N - S. So, our aim is to make the final N-S effect zero therefore we divide N - S in two equal part and it only possible when N-S is an even number hence would be "Impossible" for odd count. After dividing if possible let us say S1,S2 (equal in magnitude), all we'll have to do is make S1 - S2.

Rest i leave it upto you. And think over it when S is negative.

Solution of BLOPER spoj - spoj Operators

Maximum & Minimum Sum possible with N natural number is n*(n+1)/2 and out of it we need to take S. then we'll left with N - S. So, our aim is to make the final N-S effect zero therefore we divide N - S in two equal part and it only possible when N-S is an even number hence would be "Impossible" for odd count. After dividing if possible let us say S1,S2 (equal in magnitude), all we'll have to do is make S1 - S2.

Rest i leave it upto you. And think over it when S is negative.

#include<stdio.h> int main() { int arr[505]={0},i,j,n,s; scanf("%i%i",&n,&s); if(s < 0) { j = (n*(n+1))/2; if( (j+s)&1 || j <= -1*s ) { printf("Impossible\n"); } else { j = (j + s)/2 - 1; i = n+1; while(--i > 1 && j) { if( j - i > 1) { j -= i; arr[i] = 1; } else if(j-i == 0) { j -= i; arr[i] = 1; break; } } if(j) printf("Impossible\n"); else { printf("1"); i = 1; while(++i <= n) { arr[i] ? printf("+%i",i) : printf("-%i",i); } printf("\n"); } } } else { j = (n*(n+1))/2; if( (j-s)&1 || j < s ) { printf("Impossible\n"); } else { j = (j - s)/2; i = n+1; while(--i > 1 && j) { if( j - i > 1) { j -= i; arr[i] = 1; } else if(j-i == 0) { j -= i; arr[i] = 1; break; } } if(j) { printf("Impossible\n"); } else { printf("1"); i = 1; while(++i <= n) { arr[i] ? printf("-%i",i) : printf("+%i",i); } printf("\n"); } } } return 0; }