C语言高精度教程
- 提问者网友：轻浮
- 2021-08-11 05:23
- 五星知识达人网友：往事埋风中
- 2021-08-11 06:12
高 精 度 算 法
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <malloc.h>
int an,bn,fa=1,fb=1;
char b1[250], b2[250];
void input(int a1[],int a2[])
{
int i,ai=1,bi=1;
scanf ( "%s%s", b1, b2 );
an = strlen( b1 );
bn = strlen( b2 );
if(b1[0]==45) { an--; fa=-1;ai=0;}
if(b2[0]==45) { bn--; fb=-1;bi=0;}
for (i=0; i<an; i++,ai++) {a1[i]=b1[an-ai]-'0'; printf("%d",a1[i]);}
for (i=0; i<bn; i++,bi++) a2[i]=b2[bn-bi]-'0';
return;
}
void addition(int a[],int b[],int q)
{
int i,c[251]={0},k;
if(fa*fb>0||q)
{
if(an>bn) k=an;
else k=bn;
for(i=0;i<k;i++)
{
c[i]=a[i]+b[i]+c[i];
c[i+1]=(int)c[i]/10;
c[i]=(int)c[i]%10;
}
if(c[k]) k++;
if(fa<0&&q||fa<0) printf("-");
for(i=k-1;i>=0;i--) printf("%d",c[i]);
return;
}
return;
}
void subtraction(int a[],int b[],int q)
{
int i,f=0,c[251]={0},k;
if(fa*fb>0||q)
{
if(an>bn) k=an;
else
{ k=bn;
for(i=k;a[i]<=b[i]&&i>=0;i--)
if(a[i]<b[i]) f=1;
}
if(!f)
for(i=0;i<k;i++)
{
if(a[i]<b[i])
{ a[i+1]--;
a[i]+=10;
}
c[i]=a[i]-b[i];
}
else
for(i=0;i<k;i++)
{
if(b[i]<a[i])
{ b[i+1]--;
b[i]+=10;
}
c[i]=b[i]-a[i];
}
while(!c[k-1]&&k>1) k--;
if(q&&(fa>0&&f||fa<0&&!f)||fa>0&&(fb>0&&!f||f&&!q)) printf("-");
for(i=k-1;i>=0;i--) printf("%d",c[i]);
return;
}
}
void multiplication( int a[], int b[])
{
int i, j, c[501] = {0},k;
k = an + bn - 1;
for(i = 0; i < an; i++)
for(j = 0;j < bn; j++)
{
c[i+j] = a[i] * b[j] + c[i+j];
c[i+j+1] = c[i+j] / 10 + c[i+j+1];
c[i+j] = c[i+j] % 10;
}
while(!c[k]) k--;
if(fa*fb<0) printf("-");
for(i = k; i >= 0; i--) printf("%d",c[i]);
}
main()
{
int a[250]={0},b[250]={0};
input(a,b);
printf("\n%s+%s=",b1,b2);addition(a,b,0);
printf("\n%s-%s=",b1,b2);subtraction(a,b,0);
printf("\n%s*%s=",b1,b2);multiplication(a,b);
getchar();
}
1、 高精度除以低精度；
算法：按照从高位到低位的顺序，逐位相除。在除到第j位时，该位在接受了来自第j+1位的余数后与除数相除，如果最高位为零，则商的长度减一。源程序如下：
#include <stdio.h>
#define N 500
main()
{
int a[N] = {0}, c[N] = {0};
int i, k, d, b;
char a1[N];
printf("Input 除数:");
scanf("%d", &b);
printf("Input 被除数:");
scanf("%s", a1);
k = strlen(a1);
for(i = 0; i < k; i++) a[i] = a1[k - i - 1] - '0';
d = 0;
for(i = k - 1; i >= 0 ; i--)
{
d = d * 10 + a[i];
c[i] = d / b;
d = d % b;
}
while(c[k - 1] == 0 && k > 1) k--;
printf("商=");
for(i = k - 1; i >= 0; i--) printf("%d", c[i]);
printf("\n余数=%d", d);
}
2、高精度乘以高精度（要求用尽可能少的存储单元）；
算法：用数组保存两个高精度数，然后逐位相乘，注意考虑进位和总位数。源程序如下：
#include <stdio.h>
main()
{
int a[240] = {0}, b[240] = {0}, c[480] = {0};
int i, j, ka, kb, k;
char a1[240], b1[240];
gets(a1);
ka = strlen(a1);
gets(b1);
kb = strlen(b1);
k = ka + kb;
for(i = 0; i < ka; i++) a[i] = a1[ka-i-1] - '0';
for(i = 0; i < kb; i++) b[i] = b1[kb-i-1] - '0';
for(i = 0; i < ka; i++)
for(j = 0; j < kb; j++)
{
c[i + j] = c[i + j] + a[i] * b[j];
c[i + j +1] = c[i + j +1] + c[i + j]/10;
c[i + j] = c[i + j] % 10;
}
if(!c[k]) k--;
for(i = k-1; i >= 0; i--) printf("%d", c[i]);
}
3、高精度除以高精度（要求用尽可能少的存储单元）；
算法：用计算机模拟手算除法，把除法试商转化为连减。
#include <stdio.h>
#define N 500
int bj(int a[], int b[], int k1, int k2)
{
int i, t, flag;
if(k1 < k2)
flag = 0;
else if(k1 > k2)
flag = 1;
else
{
i = k1;
t = 0;
while(t == 0 && i > 0)
{
if(a[i] > b[i]) {t = 1; flag = 1;}
else if(a[i] == b[i]) i--;
else {t = 1; flag = 0;}
}
if(i == 0 && t == 0) flag = 2;
}
return flag;
}
int jf(int a[], int b[], int k1, int k2)
{
int i, k, d[N];
for(i = 0; i < k2; i++) d[i] = b[i];
for(i = k2; i < N; i++) d[i] = 0;
k = k1 - k2 - 1;
if(k < 0) k = 0;
if(k > 0)
{
for(i = k2 - 1; i >= 0; i--) d[i + k] = d[i];
for(i = 0; i < k; i++) d[i] = 0;
}
for(i = 0; i < k1; i++)
{
if(a[i] >= d[i]) a[i] -= d[i];
else
{
a[i + 1] = a[i + 1] - 1;
a[i] = 10 + a[i] - d[i];
}
}
return k;
}
main()
{
int a[N] = {0}, b[N] = {0}, c[N] = {0}, d[N] = {0};
int i, ka, kb, m, t, t1, t2, k, x, kd, kk;
char a1[N], b1[N];
printf("Input 被除数:");
scanf("%s", a1);
ka = strlen(a1);
for(i = 0; i < ka; i++) a[i] = a1[ka - i -1] - '0';
printf("Input 除数:");
scanf("%s", b1);
kb = strlen(b1);
for(i = 0; i < kb; i++) b[i] = b1[kb - i -1] - '0';
kd = ka;
t2 = bj(a, b, ka, kb);
m = 0;
do
{
while(a[ka - 1] == 0) ka--;
t = bj(a, b, ka, kb);
if(t >= 1)
{
k = jf(a, b, ka, kb);
c[k]++;
if(k > m) m = k;
t1 = 0;
for(i = k; i <= m; i++)
{
x = c[i] + t1;
c[i] = x % 10;
t1 = x / 10;
}
if(t1 > 0) {m++; c[m] = t1; }
}
}while(t == 1);
if(t2 == 0)
{
printf("商=0");
printf("\n余数=");
for(i = kd - 1; i >= 0; i--) printf("%d", a[i]);
exit(1);
}
if(t2 == 2)
{
printf("商 = 1");
printf("\n余数 = 0");
exit(1);
}
kk = kd;
while(!c[kd - 1]) kd--;
printf("商 = ");
for(i = kd - 1; i >= 0; i--) printf("%d", c[i]);
while(!a[kk]) kk--;
printf("\n余数 = ");
if(kk < 0)
{
printf("0");
exit(1);
}
for(i = kk; i >= 0; i--) printf("%d", a[i]);
}
4、 N！，要求精确到P位（0〈P〈1000〉。
算法：结果用数组a保存，开始时a[0]=1，依次乘以数组中各位，注意进位和数组长度的变化。源程序如下：
#include <stdio.h>
#define M 1000
main()
{
int a[M], i, n, j, flag = 1;
printf("n=");
scanf("%d",&n);
printf("n!=");
a[0] = 1;
for(i = 1; i < M; i++) a[i] = 0;
for(j = 2; j <= n; j++)
{
for(i = 0; i < flag; i++) a[i] *= j;
for(i = 0; i < flag; i++)
if(a[i] >= 10)
{
a[i+1] += a[i]/10;
a[i] = a[i] % 10;
if(i == flag-1) flag++;
}
}
for(j = flag - 1; j >= 0; j--)
printf("%d", a[j]);
}
- 1楼网友：長槍戰八方
- 2021-08-11 06:47
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- 2021-08-11 06:27