Codeforces 534C Polycarpus' Dice（思路）

C. Polycarpus' Dice

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Polycarp has n dice d1, d2, ..., dn.
The i-th dice shows numbers from 1 to di.
Polycarp rolled all the dice and the sum of numbers they showed is A. Agrippina didn't see which dice showed what number, she knows
only the sum A and the values d1, d2, ..., dn.
However, she finds it enough to make a series of statements of the
4000
following type: dice i couldn't show number r.
For example, if Polycarp had two six-faced dice and the total sum is A = 11, then Agrippina can state that each of the two dice couldn't
show a value less than five (otherwise, the remaining dice must have a value of at least seven, which is impossible).

For each dice find the number of values for which it can be guaranteed that the dice couldn't show these values if the sum of the shown values is A.

Input

The first line contains two integers n, A (1 ≤ n ≤ 2·105, n ≤ A ≤ s)
— the number of dice and the sum of shown values wheres = d1 + d2 + ... + dn.

The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 106),
where di is
the maximum value that the i-th dice can show.

Output

Print n integers b1, b2, ..., bn,
where bi is
the number of values for which it is guaranteed that the i-th dice couldn't show them.

Examples

input

2 8
44

output

3 3

input

1 3
5

output

4

input

2 3
2 3

output

0 1

Note

In the first sample from the statement A equal to 8 could be obtained in the only case when both the first and the second dice show
4. Correspondingly, both dice couldn't show values 1, 2 or 3.

In the second sample from the statement A equal to 3 could be obtained when the single dice shows 3. Correspondingly, it couldn't show
1, 2, 4or 5.

In the third sample from the statement A equal to 3 could be obtained when one dice shows 1 and the other dice shows 2. That's why
the first dice doesn't have any values it couldn't show and the second dice couldn't show 3.
题目大意：给你n个骰子，它们的点数之和为A，再给出每个骰子的最大点数，问对于每个                     骰子有多少点数是不可能出现的。
思路：不可能出现的点数等于最大点数减去可能出现的点数，所以只要求出每个骰子可出              现的最小与最大点数，用最大点数一减即可。

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;

const int INF = 1e9+10;
const int maxn = 2*1e5+10;
ll a[maxn],MIN[maxn],MAX[maxn]; //MIN[i]、MAX[i]:每个骰子的最小点数与最大点数
ll n,A,sum;

ll solve(int cur)
{
return a[cur] - (MAX[cur] - MIN[cur] + 1); //每个骰子的最大点数减去能表示的点数即其不能表示的点数
}

int main()
{
cin >> n >> A;
for(int i = 0; i < n; i++){
scanf("%I64d",&a[i]);
sum += a[i];
}
if (sum < A){
for(int i = 0; i < n; i++){
if(i) printf(" ");
printf("%I64d",a[i]);
}
printf("\n");
return 0;
}
for(int i = 0; i < n; i++){
ll yu = sum - a[i];
if (yu >= A) MIN[i] = 1; //如果减去a[i]总和还大于等于A，那么骰子i可以直接取最小值1
else MIN[i] = A - yu;   //否则骰子i的最小值就是 让其他所有骰子都取其最大值a[i],然后用A减去这些值所得的差
if (n - 1 + a[i] <= A) MAX[i] = a[i]; //如果除骰子i之外其他骰子都取1，这样的和还不大于A，那么骰子i可以取其最大值a[i]
else MAX[i] = A - (n - 1); //否则骰子i的最大值就是 让其他骰子都取1，然后用A减去这些值所得的差
}

for(int i = 0; i < n; i++){
ll ans = solve(i);
if(i) printf(" ");
printf("%I64d",ans);
}
printf("\n");
}