1 solutions
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0
C :
#include<stdio.h> #include<math.h> int main() { int a; long sums=0; scanf("%d",&a); long sum[100]; for(int i=0;i<a;i++) { if(!i) sum[i] = 2*(int)pow(10,i); else sum[i] = sum[i-1]+2*(int)pow(10,i); //printf("%ld\n",sum[i]); sums += sum[i]; } printf("%ld\n",sums); return 0; }C++ :
#include<iostream> using namespace std; int main() { long long s = 0, t = 2, n; cin >> n; while(n--) { s += t; t = t * 10 + 2; } cout << s << endl; return 0; }
- 1
Information
- ID
- 19323
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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