B. Cormen — The Best Friend Of a Man
Empirically Polycarp learned that the dog needs at least k walks for any two consecutive days in order to feel good. For example, if k = 5and yesterday Polycarp went for a walk with Cormen 2 times, today he has to go for a walk at least 3 times.
Polycarp analysed all his affairs over the next n days and made a sequence of n integers a1, a2, …, an, where ai is the number of times Polycarp will walk with the dog on the i-th day while doing all his affairs (for example, he has to go to a shop, throw out the trash, etc.).
Help Polycarp determine the minimum number of walks he needs to do additionaly in the next n days so that Cormen will feel good during all the n days. You can assume that on the day before the first day and on the day after the n-th day Polycarp will go for a walk with Cormen exactly k times.
Write a program that will find the minumum number of additional walks and the appropriate schedule — the sequence of integersb1, b2, …, bn (bi ≥ ai), where bi means the total number of walks with the dog on the i-th day.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 500) — the number of days and the minimum number of walks with Cormen for any two consecutive days.
The second line contains integers a1, a2, …, an (0 ≤ ai ≤ 500) — the number of walks with Cormen on the i-th day which Polycarp has already planned.
Output
In the first line print the smallest number of additional walks that Polycarp should do during the next n days so that Cormen will feel good during all days.
In the second line print n integers b1, b2, …, bn, where bi — the total number of walks on the i-th day according to the found solutions (ai ≤ bi for all i from 1 to n). If there are multiple solutions, print any of them.
Examples
input
3 5 2 0 1
output
4 2 3 2
input
3 1 0 0 0
output
1 0 1 0
input
4 6 2 4 3 5
output
0 2 4 3 5
题解：
题意是说有个人要遛狗，然后这条狗这要在两天内被溜一定次数就很舒服，然后现在给你这个人得日程表，表示他每天原计划溜狗多少次，然后问你还需添加多少次才能满足那条狗，给出最小值，然后输出修改后的数列
这道题很简单，我们发现如果有三天都不能满足时，我们让中间去使两边满足是最优的，那么我们就可以用贪心的思想，如果当前不行，让下一天变大。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int num[505];
int main()
{
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i++)
scanf("%d", &num[i]);
int ans = 0;
for (int i = 1; i < n; i++)
{
int ls = num[i] + num[i+1];
if (ls < m)
{
ans += m - ls;
num[i+1] += m - ls;
}
}
cout << ans << endl;
for (int i = 1; i <= n; i++)
{
cout << num[i] << " ";
}
return 0;
}