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вернуться в форумAC Solve. It's problem is very easy!! That is my AC solve. #include <stdio.h> char a[1000005], b[1000005]; int Count, min; int main() { scanf("%d", &Count); min=1; for (int i=1; i<=Count; i++) scanf("%d %d", &a[i], &b[i]); // summing for (int i=Count; i>=0; i--) { int temp = a[i] + b[i]; a[i] = temp%10; a[i-1] += temp/10; } if (a[0] == 1) min--; for (int i=min; i<=Count; i++) printf("%d", a[i]); return 0; } Re: AC Solve. It's problem is very easy!! Послано Alex 14 авг 2017 17:38 I wrote the same code on C#, but it was TLE on test#4! So, i send your code. Thank you. |
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