GYM 100090 L.Hard Problem（水~）

Description
给出前缀和数组，复原原数组
Input
第一行一整数n表示原数组长度，之后n个整数sum[i]表示前缀和(1<=n<=200000,-2e14<=sum[i]<=2e14)
Output
输出n个整数a[i]表示原数组
Sample Input
4
1 3 6 10
Sample Output
1 2 3 4
Solution
纯净水
Code

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<queue>#include<map>#include<set>#include<ctime>using namespace std;typedef long long ll;#define INF 0x3f3f3f3f#define maxn 222222int n;ll sum[maxn];int main(){    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)scanf("%I64d",&sum[i]);        sum[0]=0;        for(int i=1;i<=n;i++)            printf("%I64d%c",sum[i]-sum[i-1],i==n?'\n':' ');    }    return 0;}