Question

编写程序程序求，1-3+5-7+．．．-99=101的值

#include<iostream> using namespace std; int main() { int i = 1,sum1 = 0,j = 3,sum2 = 0,sum = 0; for(;i <= 101;i += 4) sum1 = sum1 + i; //求1+5+9+...101的值 for(;j <= 101;j += 4) sum2 = sum2 + j; //求3+7+11+...99的值 sum=sum1-sum2; cout<<"THE VALUE IS:"<<sum<<endl; return 0; }
我写的！~看看大家的~！

Answer 1

原式=(1+5+9+....+101)+(-3-7-11-....-99)
分开计算。
#include<stdio.h> /*1-3+5-7+．．．-99+101的值*/
main()
{int x1,y1,a=0,b=0,c=0;
for(x1=1;x1<=101;x1=x1+4)
a=a+x1;
for(y1=-3;y1>=-99;y1=y1-4)
b=b+y1;
c=a+b;
printf("数列和为：%d\n",c);
}

Answer 2

input n1=0,i=1
n2=0,j=3
do
n1=n1+i
i=i+4
loop until
i<101
do
n2=n2+j
j=j+4
loop until
j<=99
print s=n1-n2
end