(Problem 6)Sum square difference
?385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
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题目大意:
前十个自然数的平方和是:12?+ 22?+ ... + 102?= 385
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前十个自然数的和的平方是: (1 + 2 + ... + 10)2?= 552?= 3025
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所以平方和与和的平方的差是3025??385 = 2640.
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找出前一百个自然数的平方和与和平方的差。
#include <stdio.h>#include <string.h>#include <ctype.h>#include <math.h> #define N 100 int powplus(int n, int k){ int s=1; while(k--) { s*=n; } return s;} int sum1(int n){ return powplus((n+1)*n/2,2);} int sum2(int n){ return (n*(n+1)*(2*n+1))/6;} void solve(){ printf("%d\n",sum1(N)); printf("%d\n",sum2(N)); printf("%d\n",sum1(N)-sum2(N));} int main(){ solve(); return 0;}
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Answer:25164150