1001 Sum Problem

# 题目信息

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 745568 Accepted Submission(s): 188863

# Problem Description

Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).

In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + … + n.

# Input

The input will consist of a series of integers n, one integer per line.

# Output

For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.

# Sample Input

1
2

1
100

# Sample Output

1
2
3
4

1

5050

# 题解

# C 语言解法

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#include <stdio.h>

int main() {
    int m;
    while (EOF != scanf("%d", &m)) {
        // it not work
        // printf("%d\n\n", m*(m+1)/2);
        int s = 0;
        for (int i = 1; i <= m; i++) {
            s += i;
        }
        printf("%d\n\n", s);
    }
    return 0;
}

# Be Careful

• 输出结果后要两个换行.
• 这道题一定不要使用等差数列求和。因为 n*(n+1) 会超时 int 类型数值范围。

# 长点心吧

面前有两条路，中规中矩 and 捷径，是鼓励走捷径的。但是一定要确定捷径的结果是和另一条路是一模一样的。否则还是选择中规中矩的路吧。否则是会付出代价的。你怎么知道捷径是不是一条不归路呢？

# 链接

# 最后

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