CodeForce 534C Polycarpus' Dice （数学推理）

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 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 where s = 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.

Sample test(s)

input

2 84 4

output

3 3 

input

1 35

output

4 

input

2 32 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, 4 or 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.

题意：给出n个骰子，每个骰子有d[i]个面（点数为1~d[i]），通过摇n个骰子得到一个整数A。求每个骰子不可能出现的点数的个数，即多少个点数不可能出现。

分析：对于一个骰子，先求出其余所有筛子能够得到的最大值之和和最小值之和，然后根据最大值之和、最小值之和就能求出每个骰子不可能出现的点数。

如果其余所有的骰子都取最小值1，则最小值之和为n-1，当前骰子应取x=A - (n - 1)，当前骰子的最大取值就是x，若d[i]>x, 则从x+1~d[i]之间的数都不会取到，共有d[i]-x个；如果其余所有的骰子都去最大值d[i]，则最大值之和为sum - d[i]，当前骰子的应取y=A - (sum - d[i])，最小取值就是y，若y > 0，则从1~y-1之间的数都不可能取到，共有y-1个；二者加起来就是最终的答案。

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int N = 2e5 + 10;LL d[N];int main() {    LL n, A;    while(cin >> n >> A) {        LL sum = 0;        for(int i = 0; i < n; i++) {            cin >> d[i];            sum += d[i];        }        bool flag = false;        for(int i = 0; i < n; i++) {            LL ans = 0;            LL x = A + 1 - n;  //其它骰子都取1，当前骰子最大的取值为x            if(d[i] > x)  ans += (d[i] - x);            LL y = A - (sum - d[i]); //其它骰子都取最大值，当前骰子的最小取值为y            if(y > 0)  ans += (y - 1);            if(flag) cout << " ";            cout << ans;            flag = true;        }        cout << endl;    }    return 0;}