Sword refers to the offer series source -1+2+3+...+n

Source: Internet
Author: User

Topic 1506: Seeking 1+2+3+...+n time limit: 1 seconds Memory limit: 128 Mega Special: No submission: 1261 resolution: 723 Title Description: Ask for 1+2+3+...+n, cannot use multiplication method, for, while, if, else, switch, Case keyword and conditional judgment statement (A? B:C). Input: The input may contain multiple test samples. For each test case, the input is an integer n (1<= n<=100000). Output: Corresponds to each test case and outputs the value of the 1+2+3+...+n. Sample input: 35 Sample output: 615


Solution 1: Using constructors to simulate loops

#include <iostream> #include <stdio.h>using namespace Std;class temp{public:    Temp () {        n++;        sum+=n;    }    static void Reset () {        n=0;        sum=0;    }    static unsigned int getsum () {        return Sum;    } Private:    static unsigned int N;    static unsigned int Sum;}; unsigned int temp::n = 0;unsigned int temp::sum = 0;unsigned int Sum (int N) {    temp::reset ();    Temp::reset ();    Temp*a = new Temp[n];    Delete[] A;    A = NULL;    return Temp::getsum ();} int main () {   int n;   while (scanf ("%d", &n)!=eof) {    printf ("%d\n", SUM (n));   }    return 0;}

Solution 2: Simulating recursion with virtual functions

#include <iostream> #include <stdio.h>using namespace Std;class A; A * Array[2];class a{public:    virtual unsigned int sum (unsigned int n) {        return 0;    }}; Class B:public A{public:    virtual unsigned int sum (unsigned int n) {        return array[!! N]->sum (n-1) +n;    }; int sum (int n) {    a A;    b b;    Array[0] = &a;    ARRAY[1] = &b;    int value = Array[1]->sum (n);    return value;} int main () {   int n;   while (scanf ("%d", &n)!=eof) {    printf ("%d\n", SUM (n));   }    return 0;}


Solution 3: Using function pointers

#include <iostream> #include <stdio.h>using namespace std;typedef unsigned int (*fun) (unsigned int); unsigned int teminator (unsigned int n) {    return 0;} unsigned int sum (unsigned int n) {    static fun f[2] = {Teminator,sum};    Return n+f[!! N] (n-1);} int main () {   int n;   while (scanf ("%d", &n)!=eof) {    printf ("%d\n", SUM (n));   }    return 0;}


Sword refers to the offer series source -1+2+3+...+n

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.