Question

C语言编程求1+2+3+4+...+100。

Answer 1

代码资料：

#include "stdio.h"

int main(int argc,char *argv[]){

int s,n;

for(s=0,n=1;n<=100;s += n&1 ? -n++ : n++);

printf("-1+2-3+4-...+100 = %d\n",s);

return 0;

}

Answer 2

难嘛？
#include<stdio.h>

int main()
{
int i,flag=-1,sum=0,t;
for(i=1; i<=100; i++)
{ t = flag * i;
sum += t;
printf("%5d",t);
if(i % 5 ==0)
printf("\n");
flag = (-1)* flag;

}
printf("%d\n",sum);
}

运行结果：
-1 2 -3 4 -5
6 -7 8 -9 10
-11 12 -13 14 -15
16 -17 18 -19 20
-21 22 -23 24 -25
26 -27 28 -29 30
-31 32 -33 34 -35
36 -37 38 -39 40
-41 42 -43 44 -45
46 -47 48 -49 50
-51 52 -53 54 -55
56 -57 58 -59 60
-61 62 -63 64 -65
66 -67 68 -69 70
-71 72 -73 74 -75
76 -77 78 -79 80
-81 82 -83 84 -85
86 -87 88 -89 90
-91 92 -93 94 -95
96 -97 98 -99 100
50

Answer 3

#include <stdio.h>
int main()
{
int sum = 0, i;
for (i = 1; i <= 100; i++)
sum += i;
printf("sum=%d\n", sum);
return 0;
}