C語編程題目

① c語言編程題目

#include<stdio.h>
#include<string.h>

voidsum(char*sum,char*a,char*b)
{
	char*s_short,*s_long;
	inti_short,i_long,i_result;

	i_short=strlen(a)-1;
	i_long=strlen(b)-1;

	if(i_long>=i_short)
	{
		s_short=a;
		s_long=b;
		i_result=i_long+1;
	}
	else
	{
		s_long=a;
		s_short=b;
		i_result=i_short;
		i_short=i_long;
		i_long=i_result;
		i_result++;
	}
	for(;i_short>=0;i_short--,i_long--,i_result--)
	{
		sum[i_result]+=s_short[i_short]-'0'+s_long[i_long]-'0';
		sum[i_result-1]=sum[i_result]/10;
		sum[i_result]%=10;
		sum[i_result]+='0';
	}
	for(;i_long>=0;i_long--,i_result--)
	{
		sum[i_result]+=s_long[i_long]-'0';
		sum[i_result-1]=sum[i_result]/10;
		sum[i_result]%=10;
		sum[i_result]+='0';
	}
	if(sum[0])sum[0]+='0';
	else
	{
		i_result=strlen(sum+1);
		memmove(sum,sum+1,i_result);
		sum[i_result]=0;
	}
}
intmain(intargc,char**argv)
{
	chara[1001]={0},b[1001]={0},r[1002]={0};
	inti,j=0;
	
	scanf("%d",&i);
	while(j++<i)
	{
		memset(r,0,sizeof(r));
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		scanf("%s%s",a,b);
		
		sum(r,a,b);
		printf("Case%d:
",j);
		printf("%s+%s=%s
",a,b,r);
		if(j<i)printf("
");
	}
	return0;
}

② C語言編程題目

#include <stdio.h>
char *StrCat(char *s,char *t)
{
int i=0,j=0;
char *s1,*t1;
s1=s;t1=t;
for(;*s1;s1++,i++);
for(;*t1;t1++,j++);
if(i<=j)
{
for(;*t;)*s1++=*t++;*s1='\0';
return s;
}
else
{
for(;*s;)*t1++=*s++;*t1='\0';
return t;
}
}

int main(int argc, char *argv[])
{
char a[100],b[100];
scanf("%s",a);
scanf("%s",b);
printf("%s",StrCat(a,b));
return 0;
}

③ c語言編程題

#include<殲宏stdio.h>
void main()
{
int max,min,sum,i,score[10];
printf("輸入數據：");
for(i=0;i<判畝10;i++)
scanf("%d",&score[i]);
sum = max = min = score[0];
for(i=1; i<10; i++)
{
if(max<掘改森score[i])max = score[i];
if(min>score[i])min = score[i];
sum =sum+score[i];
}
printf("%.2f\n", (sum-min-max)/8.0);
}

④ C語言編程題，求解題思路！

只要按照公式計算每月還款金額就可以了,其中"已歸還本金累計額"等於"貸款本金 除以 還款月數 乘以 已還款月數".

按照題目要求編寫的等額本金還款的C語言程序如下

#include<stdio.h>

int main(){

double P,R,repayment;

int N,i;

scanf("%lf;%d;%lf",&P,&N,&R);

printf("[");

for(i=0;i<N;i++){

repayment=(P/N)+(P-P/N*i)*R;

if(i==N-1)

printf("%.2lf]",repayment);

else

printf("%.2lf,",repayment);

}

return 0;

}