##### solution of Save the Prisoner hackerrank Problem

A jail has N prisoners, and each prisoner has a unique id number, S, ranging from 1 to N. There are M sweets that must be distributed to the prisoners.

The jailer decides the fairest way to do this is by sitting the prisoners down in a circle (ordered by ascending S), and then, starting with some random S, distribute one candy at a time to each sequentially numbered prisoner until all M candies are distributed. For example, if the jailer picks prisoner S=2, then his distribution order would be (2,3,4,5,…,n-1, n, 1,2,3,4,…) until all M sweets are distributed.

But wait—there’s a catch—the very last sweet is poisoned! Can you find and print the ID number of the last prisoner to receive a sweet so he can be warned?

**Input Format**

The first line contains an integer, T, denoting the number of test cases.

The T subsequent lines each contain 3 space-separated integers:

N(the number of prisoners), M(the number of sweets), and S(the prisoner ID), respectively.

**Constraints**

1<=T<=100

1<=N<=10^9

1<=M<=10^9

1<=S<=10^9

**Output Format**

For each test case, print the ID number of the prisoner who receives the poisoned sweet on a new line.

**Sample Input**

1

5 2 1

**Sample Output**

2

**Explanation**

There are N=5 prisoners and S=2 sweets. Distribution starts at ID number S=1, so prisoner 1 gets the first sweet and prisoner 2 gets the second (last) sweet. Thus, we must warn prisoner 2 about the poison, so we print 2 on a new line.

How we can solve it

In this problem, we need to determine the ID of last prisoner which can be done by moving M-1 steps further from S.

ID of last prisoner= (M-1+S)

Since we are moving in a circle so we need to take mod of this summation with N.

Because the ID starts from 1, so the ID of last prisoner= (M – 1 + S – 1) % N + 1)

#### Solution in C++ of Save the Prisoner

#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int t,no,m,sw,sd; scanf("%d", &t); for(int i=0; i<t; i++) { scanf("%d %d %d", &no, &m, &sw); sd=(sw+m-1)% no; if(sd==0) printf("%d\n",no); else printf("%d\n",sd); } return 0; }

#### Solution in python

```
x=int(raw_input())
for i in range(x):
[N,M,S]=[int(j) for j in raw_input().split()]
val= (N+M+S-1)%N
if val==0:
print N
else:
print val
```

#### Solution in C

```
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int t, n, m, s;
scanf("%d", &t);
while (t--) scanf("%d%d%d", &n, &m, &s), printf("%d\n", (m+s-2)%n+1);
return 0;
}
```

#### Solution in java

```
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int rounds = input.nextInt();
for(int i = 0; i < rounds; i++)
{
int num = input.nextInt();
int lop = input.nextInt();
int s = input.nextInt() - 1;
while(lop != 0)
{
lop--;
s++;
if(s > num)
s = 1;
}
System.out.println(s);
}
}
}
```

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