B - Ticket Counter
https://atcoder.jp/contests/abc358/tasks/abc358_b
思路
第i个完成的时刻,done[i] 跟第i-1完成时间done[i-1]有关系,
第i个的开始时刻t[i] 大于 done[i-1], done[i] = t[i]+a
第i个的开始时刻t[i] 不大于 done[i-1], done[i] = done[i-1]+a
Code
https://atcoder.jp/contests/abc358/submissions/54590343
int n, a; int t[105], done[105];int main() {cin >> n >> a;for(int i=0; i<n; i++){cin >> t[i];if (i==0){done[0] = t[i]+a;} else {if (t[i] >= done[i-1]){done[i] = t[i] + a;} else{done[i] = done[i-1] + a;}}}for(int i=0; i<n; i++){cout << done[i] << endl;}return 0; }