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

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