CodeForce：732B-Cormen — The Best Friend Of a Man

传送门：http://codeforces.com/problemset/problem/732/B

Cormen — The Best Friend Of a Man

time limit per test1 second
memory limit per test256 megabytes

Problem Description

Recently a dog was bought for Polycarp. The dog’s name is Cormen. Now Polycarp has a lot of troubles. For example, Cormen likes going for a walk.

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 = 5 and 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 integers b1, 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

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解题心得：

1. 题意很简单的，就是一个狗没两天走的总和必须大于等于k，他的主人已经有了一个行走的计划，问你为了狗他必须怎么改变自己的行走计划（只能在原有的计划上加行走路程），使他多走的路最少。
2. 在比赛的时候老是读不懂题啊，慌得一批，其实就是一个贪心，每次在第二天上面加就行了。因为第二天可以作为第一天的第二天，也可以作为第三天的第二天。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1000;
int num[maxn];
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        int ans = 0;
        for(int i=2;i<=n;i++)
        {
            int sum;
            sum = num[i]+num[i-1];
            if(sum < k)
            {
                num[i] += k-sum;
                ans += k-sum;
            }
        }
        printf("%d
",ans);
        for(int i=1;i<=n;i++)
        {
            printf("%d",num[i]);
            if(i != n)
                printf(" ");
            else
                printf("
");
        }
    }
    return 0;
}