1 高精度加法
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
char s1[205],s2[205];
int sa[205],sb[205];
int len1,len2,lenn,lenm;
int ans[415];
int main(){
scanf("%s
%s",s1+1,s2+1);
len1=strlen(s1+1);
len2=strlen(s2+1);
for(int i=1,j=len1;i<=len1;i++,j--){
sa[j]=s1[i]-'0';
}
for(int i=1,j=len2;i<=len2;i++,j--){
sb[j]=s2[i]-'0';
}
lenn=len1;lenm=len2;
int lenx=max(lenn,lenm)+1;
for(int i=lenx;i>=1;i--){
ans[i]=sa[i]+sb[i];
}
for(int i=1;i<=lenx;i++){
if(ans[i]>=10){
ans[i+1]++;
ans[i]%=10;
}
}
while(ans[lenx]==0&&lenx>1) lenx--;
for(int i=lenx;i>=1;i--){
printf("%d",ans[i]);
}
return 0;
}
2 高精度减法
有符号判断
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
char num1[50005],num2[50005];
int numa1[50005],numa2[50005],ans[50005];
int len1,len2,lenn,lenm;
int main( ){
scanf("%s%s",num1+1,num2+1);
memset(ans,0,sizeof(ans));
len1=strlen(num1+1);
len2=strlen(num2+1);
lenm=max(len1,len2);
bool big=0;//big=1 :1比2大 big=0 :2比1大
if(len1!=len2) big=(len1>len2);
else{
for(int i=1;i<=len1;i++){
if(num1[i]>num2[i]){
big=1;
break;
}
if(num1[i]<num2[i]){
big=0;
break;
}
}
}
if(big==0){
swap(len1,len2);
swap(num1,num2);
}
for(int i=1,j=len1;i<=len1;i++,j--) numa1[i]=num1[j]-'0';
for(int i=1,j=len2;i<=len2;i++,j--) numa2[i]=num2[j]-'0';
for(int i=1;i<=lenm;i++){
ans[i]=numa1[i]-numa2[i];
}
for(int i=1;i<=lenm;i++){
if(ans[i]<0){
ans[i]+=10;
ans[i+1]--;
}
else continue;
}
lenn=lenm;
while(ans[lenn]==0&&lenn>=0){
lenn--;
if(lenn==-1) {
printf("0");
return 0;
}
}
if(big==0) printf("-");
for(int i=lenn;i>=1;i--){
printf("%d",ans[i]);
}
return 0;
}
3 高精度除以低精度
A 本人写法
//本人写法
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
char num1[205];
int ans[205],numa1[205],numa2[205];
int len1,len2,lenm;
int ys;//余数
int ss;//商
int f=0;
long long x;
int main( ){
scanf("%s",num1+1);
len1=strlen(num1+1);
scanf("%lld",&x);
if(x==0) {
printf("0");
return 0;
}
for(int i=1;i<=len1;i++)
{
numa1[i]=num1[i]-'0';
}
for(int i=1;i<=len1;i++){
ys=numa1[i]%x;
ss=numa1[i]/x;
numa1[i]=ss;
numa1[i+1]+=ys*10;
}
for(int i=1;i<=len1;i++){
if(f==0&&numa1[i]==0) printf("0");
if(f==0&&numa1[i]>0) f=1;
if(f==1){
printf("%d",numa1[i]);
}
}
return 0;
}
B 洛谷题解其一
#include<cstdio>
#include <cstring>
using namespace std;
int a[10000];
int y;
int l=0;
int yushu;
int shang;
int f=0;
int main()
{
char c=getchar();
while (c>='0' && c<='9')
{
l++;
a[l]=(c-'0');
c=getchar();
}
//scanf("%s",a);
//l=strlen(a);
scanf("%d",&y);
for (int i=1;i<=l;i++)
{
yushu=a[i]%y;//第i位的余数
shang=a[i]/y;//第i位的商
a[i]=shang;
a[i+1]+=yushu*10;//把余数弄到下一位
}
for (int i=1;i<=l;i++)
{
if (f==0 && a[i]>0) f=1;
if (f==1) printf("%d",a[i]);
}
return 0;
}
4 高精度乘法
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
char num1[2005],num2[2005];
int numa1[2005],numa2[2005],ans[4005];
int len1,len2,lenn,lenm;
int lenn1,lenn2;
int main( ){
scanf("%s%s",num1+1,num2+1);
len1=strlen(num1+1);
len2=strlen(num2+1);
lenn1=len1;
lenn2=len2;
memset(numa1,0,sizeof(numa1));
memset(numa2,0,sizeof(numa2));
memset(ans,0,sizeof(ans));
for(int i=1,j=len1;i<=len1;i++,j--) numa1[i]=num1[j]-'0';
for(int i=1,j=len2;i<=len2;i++,j--) numa2[i]=num2[j]-'0';
//for(int i=lenn1;i>=1;i--){
// while(numa1[i]==0) len1--;
//}
//for(int i=lenn2;i>=1;i--){
// while(numa2[i]==0) len2--;
//}
//前导0考虑 //判断0的情况
for(int i=1;i<=len1;i++){
for(int j=1;j<=len2;j++){
ans[i+j-1]+=numa1[i]*numa2[j];
if(ans[i+j-1]>=10){
ans[i+j]+=ans[i+j-1]/10;
ans[i+j-1]%=10;
}
}
}
lenn=len1+len2;
for(int i=1;i<=lenn;i++){
if(ans[i]>=10){
ans[i+1]+=ans[i]/10;
ans[i]%=10;
}
}
while(ans[lenn]==0&&lenn>1){
lenn--;
}
for(int i=lenn;i>=1;i--){
printf("%d",ans[i]);
}
return 0;
}
5 高精度(2^N)
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <iomanip>
using namespace std;
int ans[1005]={0};
void doubl2(int c[]){
for(int i=1;i<=c[0];i++){
c[i]=c[i]*2;
}
for(int i=1;i<=c[0];i++){
if(c[i]>=10){
c[i+1]+=ans[i]/10;//ans??
c[i]%=10;
}
}
if(c[c[0]+1]) c[0]++;
}
void prta(int c[]){
for(int i=1;i<=c[0];i++){
printf("%d",c[c[0]+1-i]);
}
}
int main( ){
ans[1]=1;
ans[0]=1;
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++){
doubl2(ans);
}
prta(ans);
return 0;
}
6 高精度合集
By _cyl
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=2e4+5;
char s[N];
ll qwq,MOD;
struct bignum
{
int a[N],len;
bignum(){memset(a,0,sizeof(a));len=0;}
void push()
{
for(int i=0;i<=len;i++)
{
if(a[i]>=10)
{
if(i==len) len++;
a[i+1]+=a[i]/10;
a[i]%=10;
}
}
}
void read()
{
scanf("%s",s);
len=strlen(s)-1;
for(int i=0,j=len;i<=len;i++,j--) a[i]=s[j]-'0';
}
void write()
{
for(int i=len;i>=0;i--) printf("%d",a[i]);
putchar('
');
}
friend inline bool operator <(const bignum &x,const bignum &y)
{
if(x.len!=y.len) return x.len<y.len;
for(int i=x.len;i>=0;i--) if(x.a[i]!=y.a[i]) return x.a[i]<y.a[i];
return 0;
}
friend inline bignum operator +(const bignum &x,const bignum &y)
{
bignum ans;
ans.len=max(x.len,y.len);
for(int i=0;i<=ans.len;i++) ans.a[i]=x.a[i]+y.a[i];
ans.push();
return ans;
}
friend inline bignum operator -(const bignum &fi,const bignum &se)
{
bignum ans,x,y;
int op=1;
if(fi<se) op=-1,x=se,y=fi;
else
{
if(!(se<fi)) return ans;
x=fi,y=se;
}
//x.write();y.write();
ans.len=x.len;
for(int i=0;i<=ans.len;i++)
{
ans.a[i]=x.a[i]-y.a[i];
if(ans.a[i]<0) ans.a[i]+=10,x.a[i+1]--;
}
for(int i=ans.len;i>=0;i--) if(ans.a[i]==0) ans.len--;else break;
ans.a[ans.len]*=op;
return ans;
}
friend inline bignum operator *(const bignum &x,const bignum &y)
{
bignum ans;
if((x.len==0 && x.a[0]==0) || (y.len==0 && y.a[0]==0))return ans;
ans.len=x.len+y.len;
for(int i=0;i<=x.len;i++)
for(int j=0;j<=y.len;j++)
ans.a[i+j]+=x.a[i]*y.a[j];
ans.push();
return ans;
}
friend inline bignum operator /(const bignum &x,const ll &y)
{
ll alpha=0;
bignum ans;
for(int i=x.len;i>=0;i--)
{
alpha=alpha*10+x.a[i];
if(alpha>=y && ans.len==0) ans.len=i;
ans.a[i]=alpha/y;alpha%=y;
}
MOD=alpha;
return ans;
}
}f[2];
int main()
{
//f[0].read();//f[0].write();
//scanf("%lld",&qwq);//f[1].write();
//(f[0]/qwq).write();
//printf("%lld
",MOD);
f[0].read();f[1].read();(f[0]-f[1]).write();
return 0;
}