codeforces 732B - Cormen --- The Best Friend Of a Man

B. Cormen — The Best Friend Of a Man

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 = 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.

简单的贪心算法，就是当两个相加小于k时，选择将后面一个数增大；

AC代码：

 

# include <stdio.h>
using namespace std;
typedef long long int ll;
int ans[510], a[510];
int main(){
	int i,j, k, n;
	scanf("%d%d", &n, &k);
	for(i=1; i<=n; i++){
		scanf("%d", &a[i]);
	}
	if(n==1){
		printf("%d\n%d", 0, a[1]);
	}
	else{
		ans[1]=a[1];
		int sum=0;
		for(i=2; i<=n; i++){
			if(a[i]+ans[i-1]<k){
				sum=sum+k-a[i]-ans[i-1];
				ans[i]=k-ans[i-1];
			}
			else{
				ans[i]=a[i];
			}
		}
		printf("%d\n", sum);
		for(i=1; i<=n; i++){
			if(i!=n){
				printf("%d ", ans[i]);
			}
			else{
				printf("%d", ans[i]);
			}
		}
	}
	return 0;
}