Henu ACM Round#21 B. inc ARG

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B. inc ARG

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of n bits. These bits are numbered from 1 to n. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the n-th bit.

Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.

Sergey wrote certain values ​​of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of bits in the cell.

The second line contains a string consisting of n characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.

Output

Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.

Examples

input

4
1100

output

3

input

4
1111

output

4

Note

In the first sample the cell ends up with value 0010, in the second sample — with 0000.

题意：给一个100位以内的二进制的数，加1后有多少位发生变化。

解法：这个二进制从左往右是最低位--最高位，跟一般的从左到右是最高位到最低位一样，并没有什么差别，只是加1，从左边第个数开始加，进位往后面一位进位

代码：

方法一：计算加1之后的二进制串，跟原来的二进制串对比

import java.util.Scanner;

public class Main {

	static final int maxn = 100+10;
	static int[] arr = new int[maxn];
	static int[] res = new int[maxn];
	
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n;
		String str;
		while(scanner.hasNext()){
			n = scanner.nextInt();
			str = scanner.next();
			int bit = 1;
			for(int i=0;i

方法二：任何一位只要加1则改变，直接统计二进制串中有多少位加1了（第一位加1是题上要求，其他位加1则是因为进位），所以不用存加1的结果（如果是c或c++，处理字符串直接str[index]即可，不用像java这么麻烦）

import java.util.Scanner;

public class Main1 {

	static final int maxn = 100 + 10;
	static char[] arr = new char[maxn];

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n;
		String str;
		while (scanner.hasNext()) {
			n = scanner.nextInt();
			str = scanner.next();
			for (int i = 0; i < str.length(); i++) {
				arr[i] = str.charAt(i);
			}
			arr[0] += 1;
			int ans = 0;
			for (int i = 0; i < n; i++) {
				if (arr[i] == '2') {
					ans++;
					if (i + 1 == n)
						break;
					arr[i + 1] += 1;
				} else {
					ans++;
					break;
				}
			}
			System.out.println(ans);
		}
	}
}