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 integers b1, b2, ..., bn (bi ≥ ai), where bi means the total number of walks with the dog on the i-th day.
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.
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.
3 5
2 0 1
4
2 3 2
3 1
0 0 0
1
0 1 0
4 6
2 4 3 5
0
2 4 3 5
第一行输入天数n和两天最少步行数k,第二行输入计划步行数,要求输出最少增加的步数和每天的步数,题目实在太长,很长时间才搞明白这道题要做啥。解决方法是优先把步数增加在第二天,因为增加到第二天后在后一天的计算中会减少步数的增加。
#include<iostream> #define N 505 using namespace std; int main() { int n, k, walk[N], minwalk=0, twodayWalk; cin >> n >> k; for (int i = 0;i < n;i++) cin >> walk[i]; for (int i = 1;i < n;i++) { twodayWalk = walk[i] + walk[i - 1]; if (twodayWalk < k) { walk[i] = k - walk[i - 1]; minwalk += k - twodayWalk; } else continue; } cout << minwalk << endl; for (int i = 0;i < n;i++) cout << walk[i] << ' '; cout << endl; return 0; }