ny 436 sum of all integer numbers
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#include<stdio.h>#include<stdlib.h>int main(){ int n; while(scanf("%d",&n)!=EOF) { printf("%d\n",(abs(n-1)+1)*(n+1)/2); }return 0; }
原题链接
错了就好好思考错在哪里了,编程要考虑全面
这题主要是负数和等差数列
等差数列求和公式s=(a1+an)*n/2=a1*n+n*(n-1)*d/2;
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