题目要求是这样的,计算两个数的乘积
试一下我的这个吧
#include <stdio.h>
int main() {
int a, b;
printf("请输入小灯的总行数和总列数(以空格分隔):");
scanf("%d %d", &a, &b);
if (a >= 1 && a <= 100000 && b >= 1 && b <= 100000) {
// 计算小灯的总数
long long int total = (long long int)a * b;
printf("小灯的总数为:%lld\n", total);
} else {
printf("输入的行数和列数超出允许的范围。\n");
}
return 0;
}
超出int的表示范围,溢出了呀
int改成long long
%d改成%lld
把所有的 int 改成 long long,然后把 printf("%d")那里改成 %lld。
大概率是溢出了。
100000 * 100000 的结果超出 int 型数据范围了,溢出了,结果就不正确了。修改如下,供参考:
#include <stdio.h>
int main()
{
long long a, b;
do{
scanf("%lld %lld", &a, &b);
if (a < 1 || a > 100000 || b < 1 || b > 100000)
printf("请重新输入两个整数:");
else
break;
} while (1);
printf("%lld", a * b);
return 0;
}
硬件连接:
代码:
#include <reg52.h>
#include <intrins.h>
#define uchar unsigned char
#define uint unsigned int
uchar Count;
sbit Dot = P0^7;
uchar code DSY_CODE[]=
{
0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f
};
uchar Digits_of_6DSY[]={0,0,0,0,0,0};
void DelayMS(uint x)
{
uchar i;
while(--x)
{
for(i=0;i<120;i++);
}
}
void main()
{
uchar i,j;
P0 = 0x00;
P3 = 0xff;
Count =0;
TMOD = 0x01;
TH0 = (65535-50000)/256;
TL0 = (65535-50000)%256;
IE = 0x82;
TR0 = 1;
while(1)
{
j = 0x7f;
for(i=5;i!=-1;i--)
{
j=_crol_(j,1);
P3 = j;
P0 = DSY_CODE[Digits_of_6DSY[i]];
if(i==1) P0 |= 0x80;
DelayMS(2);
}
}
}
void Time0() interrupt 1
{
uchar i;
TH0 = (65535-50000)/256;
TL0 = (65535-50000)%256;
if(++Count !=2) return;
Count = 0;
Digits_of_6DSY[0]++;
for(i=0;i<=5;i++)
{
if(Digits_of_6DSY[i] == 10)
{
Digits_of_6DSY[i] = 0;
if(i != 5) Digits_of_6DSY[i+1]++;
}
else break;
}
}