【56.74%】【codeforces 732B】Cormen --- The Best Friend Of a Man

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 = 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

【题解】

只要前一天和今天的次数和没有大于等于k，则增加当前天的次数，直至前一天和今天的次数和等于k。
勇敢贪吧。

#include <cstdio>int a[600];int n, k;int main(){    //freopen("F:\\rush.txt", "r", stdin);    scanf("%d%d", &n, &k);    for (int i = 1; i <= n; i++)        scanf("%d", &a[i]);    a[0] = k; a[n + 1] = k;    int add = 0;    for (int i = 1; i <= n; i++)    {        int num = a[i - 1] + a[i];        if (num < k)        {            add += (k - num);            a[i] += (k - num);        }    }    printf("%d\n", add);    for (int i = 1; i <= n; i++)        printf("%d%c", a[i], i == n ? '\n' : ' ');    return 0;}