第一个只出现一次的字符，josephus环，最大子数组和

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAXINT 0x7fffffff
#define MININT 0X80000000
//字符串中第一个只出现一次的字符
char firstSingle(char *str)
{
	int a[255];
	memset(a, 0, 255 * sizeof(int));
	char *p = str;
	while (*p != ‘\0‘){
		++a[*p];
		++p;
	}
	p = str;
	while (*p != ‘\0‘){
		if (a[*p] == 1) return *p;
		++p;
	}
	return ‘\0‘;
}
//josephus问题 f(n,m)=(f(n-1,m)+m)%n,迭代求解最后一个数的原始数
int joseph(int n, int m)
{
	int fn = 0;
	for (int i = 2; i <= n; ++i)
		fn = (fn + m) % i;
	return fn;
}

//最大子数组和
int maxSubArraySum(int *a, int n, int &b, int &e)
{
	int p, sum, maxSum;
	p = sum = 0;
	maxSum = MININT;
	for (int i = 0; i < n; ++i){
		if (sum < 0){
			sum = a[i];
			p = i;
		}
		else sum += a[i];
		if (sum > maxSum){
			maxSum = sum;
			b = p;
			e = i;
		}
	}
	return maxSum;
}
int main()
{
	char str[] = "abaccdeff";
	printf("%c\n", firstSingle(str));
	printf("%d\n", joseph(5, 3));
	int a[] = { 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };
	int len, b, e;
	len = maxSubArraySum(a, 16, b, e);
	printf("the maxSubArray is :\n");
	for (int i = b; i <= e; ++i)
		printf("%d\t", a[i]);
	printf("\nthe maxSubArraySum is : %d\n", len);
}