[C++知识库]《C语言入门100例》(第2例) 求和

今天的打卡章节为《C语言入门100例》(第2例) 求和；

附上链接：【第02题】给定 n，求 1 + 2 + 3 + ... + n 的和 | 四种解法

全文目录

主要内容

? ? ?求和方法：

力扣习题：

????????剑指 Offer 64. 求1+2+…+n

????????剑指 Offer 57 - II. 和为s的连续正数序列

主要内容? ? ?

????????求和方法：

? ? ? ? 解法一：循环枚举

#include <stdio.h>
int main() {
    int n, ans;
    while (scanf("%d", &n) != EOF) {
        ans = 0;          
        while(n) {        
            ans += n;     
            --n;             
         }
        printf("%d\n\n", ans);
    }
    return 0;
}

? ? ? ? 解法二：奇偶性判断

#include <stdio.h>
int main() {
    int n, ans;
    while (scanf("%d", &n) != EOF) {
        if(n % 2 == 0) {           
            ans = n / 2 * (n+1);   
        } else {                   
            ans = (n+1) / 2 * n;   
        }
        printf("%d\n\n", ans);
    }
    return 0;
}

? ? ? ? 解法三：无符号整型

#include <stdio.h>
int main() {
    unsigned int n;
    while (scanf("%u", &n) != EOF) {
        unsigned int ans = n * (n + 1) / 2; 
        printf("%u\n\n", ans);
    }
    return 0;
}

? ? ? ? 解法四：64位整型

#include <stdio.h>
int main() {
    long long n;
    while (scanf("%lld", &n) != EOF) {
        long long ans = n * (n + 1) / 2;  
        printf("%lld\n\n", ans);
    }
    return 0;
}

力扣习题：

????????剑指 Offer 64. 求1+2+…+n

直接递归相加

int sumNums(int n){
    if(n==0)
    {
        return 0;
    }
    return n+sumNums(n-1);
}

????????剑指 Offer 57 - II. 和为s的连续正数序列

n的和只能在n/2 + 1之前产生，利用双指针遍历1到n/2查找符合的条件，具体看代码。

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** findContinuousSequence(int target, int* returnSize, int** returnColumnSizes) {
    if (target < 1) {
        return NULL;
    }
    int** ans = (int**)malloc(sizeof(int*) * target);
    *returnColumnSizes = (int*)malloc(sizeof(int) * target);
    int left = 1, right = 1, sum = 0, len = target / 2;
    *returnSize = 0;
    while (left <= len) {
        if (sum == target) {
            (*returnColumnSizes)[*returnSize] = right - left;
            ans[*returnSize] = (int*)malloc(sizeof(int) * (right - left));
            for (int i = left; i < right; i++) {
                ans[*returnSize][i - left] = i;
            }
            (*returnSize)++;
            sum -= left++;
        }
        if (sum < target) {
            sum += right++;
        }
        if (sum > target){
            sum -= left++;
        }
    }
    return ans;
}