【codeforces 723 B Cormen — The Best Friend Of a Man】

Cormen — The Best Friend Of a Man
B. 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

思路还是很好想的 : 在当前啊a[i] + a[i + 1] < k 时，a[i + 1] += k - a[i] ,需要多走的步数 ans += k - a[i ];

#include<cstdio>int pa[500];int main(){   int N,M,ans = 0,i,cut;   scanf("%d %d",&N,&M);   for(i = 1 ; i <= N ; i++)       scanf("%d",&pa[i]);   for(i = 2 ; i <= N; i++){       cut = pa[i - 1] + pa[i];       if(cut < M){           pa[i] += M - cut;           ans += M - cut;       }   }   printf("%d\n",ans);   for(i = 1 ; i < N; i++)       printf("%d ",pa[i]);   printf("%d\n",pa[N]);   return 0;}