CodeForces 732B Cormen — The Best Friend Of a Man
1 second
256 megabytes
standard input
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.
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
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h> using namespace std;
int n,k;
int a[505];
int b[505];
int main()
{
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
int ans=0;
for(int i=2;i<=n;i++)
{
if(a[i]+a[i-1]>=k) continue;
else
{
ans+=(k-a[i]-a[i-1]);
a[i]+=(k-a[i]-a[i-1]);
} }
printf("%d\n",ans);
for(int i=1;i<=n;i++)
{
if(i!=n)
printf("%d ",a[i]);
else
printf("%d\n",a[i]);
}
return 0; }
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