压八位高精度 高精操作大全
1.高精简单操作
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
struct bigint{
static const int base=1e+8;
static const int width=8;
int s[10001],tot;
bigint(long long num=0){*this=num;}
void lenth(){while(tot>0&&!s[tot-1]) --tot;}
bigint operator = (long long num){//
tot=0;
while(num){
s[tot++]=num%base;
num/=base;
}
this->lenth();
return *this;
}
bigint operator = (char* buf){//
tot=0;
int len=strlen(buf),x=0;
for(register int i=len-1;i>=0;i-=width){
x=0;
for(register int j=max(i-width+1,0);j<=i;++j) x=x*10+buf[j]-'0';
s[tot++]=x;
}
this->lenth();
return *this;
}
int& operator [] (int x){//
return s[x];
}
void print(){//
lenth();
if(!tot){
putchar('0');
return;
}
int p=tot-1;
printf("%d",s[p]);
for(p=tot-2;p>=0;--p) printf("%08d",s[p]);
}
bigint operator + (bigint b){//
bigint c=0;
int x=0;
for(register int i=0;i<tot||i<b.tot;++i){
if(i<tot) x+=s[i];
if(i<b.tot) x+=b[i];
c[c.tot++]=x%base;
x/=base;
}
while(x){
c[c.tot++]=x%base;
x/=base;
}
c.lenth();
return c;
}
friend bool operator < (bigint a,bigint b){//
if(a.tot!=b.tot) return a.tot<b.tot;
for(register int i=a.tot-1;i>=0;--i){
if(a[i]!=b[i]) return a[i]<b[i];
}
return false;
}
friend bool operator == (bigint a,bigint b){//
a.lenth();
b.lenth();
return (!(a<b)&&!(b<a));
}
bool pd(){//
return !(s[0]&1);
}
bigint operator - (bigint b){//
bigint c=0;
int x=0,g=0;
for(register int i=0;i<tot||i<b.tot||g<0;++i){//////////////////////////////////////////
x=g;
g=0;
if(i<tot) x+=s[i];
if(i<b.tot) x-=b[i];
while(x<0){
x+=base;
--g;
}
c[c.tot++]=x%base;
g+=x/base;
}
c.lenth();
return c;
}
bigint divide(){//
bigint c=0;
long long int x=0;
c.tot=tot;
for(register int i=tot-1;i>=0;--i){
c[i]=(x*base+s[i])>>1;
x=(s[i]&1);
}
c.lenth();/////////////////////////////
return c;
}
bigint multi(){//
long long x=0;
bigint c=0;
for(register int i=0;i<tot;++i){
x+=(s[i]<<1);
c[c.tot++]=x%base;
x/=base;
}
while(x){
c[c.tot++]=x%base;
x/=base;
}
return c;
}
bigint operator << (int x){//
bigint c=(*this);
while(x--) c=c.multi();
c.lenth();
return c;
}
};
const bigint cmp=0;
bigint gcd(bigint a,bigint b){
int mul=0;
bigint c;
while(1){
if(b==cmp) return mul==0?(a):(a<<mul);
while((a.pd())&&(b.pd())){
a=a.divide();
b=b.divide();
++mul;
}
while(a.pd()) a=a.divide();
while(b.pd()) b=b.divide();
if(a<b) swap(a,b);
c=a-b;
if(c<b){
a=b;
b=c;
}
else a=c;
}
}
int main(){
char st[10010];
bigint a,b,c;
scanf("%s",st);
a=st;
scanf("%s",st);
b=st;
if(a<b) swap(a,b);
c=gcd(a,b);
c.print();
return 0;
}
高精，GCD，stein.
即
gcd(a,b)=2*gcd(a/2,b/2);
gcd(a,b)=gcd(a/2,b);
gcd(a,b)=gcd(a,b/2);
再用辗转相减套个高精就好
2.高精除int
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int num[10001],l1=0;
int res[10001],len=0;
struct bigint{
static const int base=10;
static const int width=1;
int tot;
int s[10001];
bigint(long long num=0){(*this)=num;}
void lenth(){while(tot>0&&!s[tot-1]) --tot;}
int& operator [] (int n){
return s[n];
}
bigint operator = (long long num){
tot=0;
while(num){
s[tot++]=num%base;
num/=base;
}
this->lenth();
return *this;
}
bigint operator = (char* buf){
tot=0;
int len=strlen(buf);
for(register int i=len-1;i>=0;--i){
s[tot++]=buf[i]-'0';
}
this->lenth();
return *this;
}
bigint mul(int n){
long long x=0;
bigint c=0;
for(register int i=0;i<tot;++i){
x+=s[i]*n;
c[c.tot++]=x%base;
x/=base;
}
while(x){
c[c.tot++]=x%base;
x/=base;
}
return c;
}
void print(){
lenth();
if(!tot){
putchar('0');
return;
}
int p=tot-1;
printf("%d",s[p]);
for(p=tot-2;p>=0;--p) printf("%d",s[p]);
return;
}
bigint div(int n){
if(n==1) return *this;
bigint c;
long long x=0;
/*for(register int i=tot-1;i>=0;--i){
if((s[i]+(x<<1)+(x<<3))>=n){
x=(x<<1)+(x<<3)+s[i];
c[i]=x/n;
c.tot++;
x%=n;
}
else{
x=(x<<1)+(x<<3)+s[i];
}
}*/
for(register int i=tot-1;i>=0;--i){
x=(x<<1)+(x<<3)+s[i];
c[i]=x/n;
x%=n;
}
c.tot=tot;
c.lenth();
return c;
}
};
bigint katalan(int n){
bigint c=1;
int N=(n<<1);
for(register int i=n+2;i<=N;++i)
c=c.mul(i);
for(register int i=1;i<=n;++i)
c=(c.div(i));
return c;
}
int main(){
int n;
bigint a;
scanf("%d",&n);
a=katalan(n);
a.print();
return 0;
}
高精，katalan，组合数学.
即
暴力打表观察分析一下，易得答案为katalan数
递推显然是要T的，化简一波就好了
再套个高精，支持除非高精就行
3.高精乘高精
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
struct bigint{
static const int base=10;
static const int width=1;
int tot,s[10005];
bigint(long long num=0){*this=num;memset(s,0,sizeof(s));tot=0;}/////////////memset?????
void lenth(){while(tot>0&&!s[tot-1])--tot;}
bigint operator = (long long num){
tot=0;
while(num){
s[tot++]=num%base;
num/=base;
}
this->lenth();
return *this;
}
/*bigint operator = (char* buf){
tot=0;
int len=strlen(buf),x=0;
for(register int i=len-1;i>=0;i-=width){
x=0;
for(register int j=max(i-width+1,0);j<=i;++j) x=(x<<1)+(x<<3)+(buf[j]^48);
s[tot++]=x;
}
this->lenth();
return *this;
} */
bigint operator = (string buf){
tot=0;
int len=buf.length(),x=0;
for(register int i=len-1;i>=0;i--){
x=0;
for(register int j=max(i-width+1,0);j<=i;++j) x=(x<<1)+(x<<3)+(buf[j]^48);
s[tot++]=x;
}
this->lenth();
return *this;
}
int& operator [] (int x){
return s[x];
}
friend bool operator < (bigint a,bigint b){
if(a.tot!=b.tot) return a.tot<b.tot;
for(register int i=a.tot-1;i>=0;--i){
if(a[i]!=b[i]) return a[i]<b[i];
}
return false;
}
void print(){
lenth();
if(!tot){
putchar('0');
return;
}
for(int p=tot-1;p>=0;--p) printf("%d",s[p]);
return;
}
void carry_up(){
int rep=tot-1;
for(register int i=0;i<rep;++i){
if(s[i]>=base){
s[i+1]+=s[i]/base;
s[i]%=base;
}
}
int estbit=tot-1,res=s[tot-1]/base;
while(res){
++estbit;
res/=base;
}
//printf("the highest bit of the array is %d\n",estbit);//DEBUG
for(register int i=tot-1;i<=estbit;++i){
if(s[i]>=base){
s[i+1]=s[i]/base;
s[i]%=base;
}
}
tot+=estbit+1;
this->lenth();
return;
}
bigint multiply(bigint b){
bigint c=0;
c.tot=0;
if(tot>b.tot){
//printf("a.tot>b.tot\n"); // DEBUG
for(register int i=0;i<b.tot;i++){
for(register int j=0;j<tot;j++){
//c[i]=s[i]*b[i];
c[i+j]+=b[i]*s[j];
}
}
}
else{
//printf("a.tot<=b.tot\n"); //DEBUG
for(register int i=0;i<tot;i++){
for(register int j=0;j<b.tot;j++){
//c[i]=s[i]*b[i];
c[i+j]+=s[i]*b[j];
}
}
}
c.tot=tot+b.tot;
c.lenth();
//printf("c.tot=%d\n",c.tot);//DEBUG
c.carry_up();
return c;
}
};
bigint num[41][41],f[41][8];
int n,k;
char str[42];
void getnums(){
string s;//cout<<n<<endl;
for(register int i=1;i<=n;++i){
s.clear();
int j=i;
for( ;j<=n;++j){
s=s+str[j-1];
//cout<<s<<endl;//DEBUG
num[i][j]=s;
}
num[i][j].tot=s.length();
}
return;
}
inline bigint max_bigint(bigint a,bigint b){
if(a<b) return b;
return a;
}
int main(){
scanf("%d%d",&n,&k);
scanf("%s",str);
/*for(register int i=1;i<=n;++i){
for(register int j=i;j<=n;++j){
for(register int l=i;l<=j;++l){
num[i][j]=num[i][j]*10+(ar[l-1]^48);
}
}
}*/
getnums();
/*for(register int i=1;i<=n;++i){//DEBUG
for(register int j=i;j<=n;++j){
num[i][j].print();
putchar(' ');
}
putchar('\n');
}*/
for(register int i=1;i<=n;++i){
f[i][0]=num[1][i];
}
for(register int i=1;i<=n;++i){
for(register int j=1;j<=k;++j){
for(register int l=j;l<=i;++l){
/*f[l][j-1].print();
printf("(f[%d][%d]) ",l,j-1);
putchar('*');
num[l+1][i].print();
printf("(num[%d][%d]) ",l+1,i);
putchar('=');
bigint mid;
mid=(f[l][j-1]).multiply(num[l+1][i]);
mid.print();
printf("(tot=%d) ",mid.tot);
putchar('\n');
//DEBUG
printf("f[%d][%d]=max(",i,j);//DEBUG
f[i][j].print();
putchar(',');
mid.print();
putchar(')');
f[i][j]=max_bigint(f[i][j],mid);
putchar('=');
f[i][j].print();
putchar('\n');
putchar('\n');*/
f[i][j]=max_bigint(f[i][j],f[l][j-1].multiply(num[l+1][i]));
}
}
}
/*for(register int i=1;i<=n;++i){//DEBUG
for(register int j=i;j<=n;++j){
printf("%lld ",num[i][j]);
}
putchar('\n');
}*/
f[n][k].print();
return 0;
}
高精，DP.
DP不难，高精乘高精倒是DEBUG了一晚上...
4.高精大全
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
struct bigint{
static const int base=1e+8;
static const int width=8;
int tot;
long long s[10005];
bigint(long long num=0){*this=num;memset(s,0,10005);tot=0;}
void lenth(){while(tot>0&&!s[tot-1]) --tot;}
bigint operator = (long long num){
tot=0;
while(num){
s[tot++]=num%base;
num/=base;
}
this->lenth();
return *this;
}
bigint operator = (char* buf){
tot=0;
int len=strlen(buf),x=0;
for(int i=len-1;i>=0;i-=width){
x=0;
for(int j=max(i-width+1,0);j<=i;++j)
x=x*10+buf[j]-48;
s[tot++]=x;
}
this->lenth();
return *this;
}
long long int& operator [](int x){
return s[x];
}
void print(){
lenth();
if(!tot){
putchar('0');
return;
}
int p=tot-1;
printf("%lld",s[p]);
for(p=tot-2;p>=0;p--)
printf("%08lld",s[p]);
return;
}
friend bool operator < (bigint a,bigint b){
if(a.tot!=b.tot) return a.tot<b.tot;
for(int i=a.tot-1;i>=0;--i)
if(a[i]!=b[i]) return a[i]<b[i];
return false;
}
friend bool operator == (bigint a,bigint b){
a.lenth();
b.lenth();
return ((!(a<b))&&(!(b<a)));
}
friend bool operator <=(bigint a,bigint b){
return ((a<b)||(a==b));
}
bigint operator + (bigint b){
bigint c=0;
int x=0;
for(int i=0;i<tot||i<b.tot;++i){
(i<tot)?(x+=s[i]):1;
(i<b.tot)?(x+=b[i]):1;
c[c.tot++]=x%base;
x/=base;
}
while(x){
c[c.tot++]=x%base;
x/=base;
}
c.lenth();
return c;
}
bigint operator - (bigint b){
bigint c=0;
int x=0,g=0;
for(int i=0;i<tot||i<b.tot||g<0;++i){
x=g;
g=0;
(i<tot)?(x+=s[i]):1;
(x<b.tot)?(x-=b[i]):1;
while(x<0){
x+=base;
--g;
}
c[c.tot++]=x%base;
x/=base;
}
c.lenth();
return c;
}
void carry_up(){
int rep=tot-1;
for(int i=0;i<rep;i++){
(s[i]>=base)?(s[i+1]+=s[i]/base,s[i]%=base):1;
}
int estbit=rep,res=s[rep]/base;
while(res){
++estbit;
res/=base;
}
for(int i=tot-1;i<=estbit;i++){
(s[i]>=base)?(s[i+1]+=s[i]/base,s[i]%=base):1;
}
tot+=estbit+1;
this->lenth();
return;
}
bigint multiply(bigint b){
bigint c=0;
c.tot=0;
if(tot>b.tot){
for(int i=0;i<b.tot;i++){
for(int j=0;j<tot;j++){
c[i+j]+=b[i]*s[j];
}
}
}
else{
for(int i=0;i<tot;i++){
for(int j=0;j<b.tot;j++){
c[i+j]+=s[i]*b[j];
}
}
}
c.tot=tot+b.tot;
c.lenth();
c.carry_up();
return c;
}
bigint divide(){//
bigint c=0;
long long int x=0;
c.tot=tot;
for(register int i=tot-1;i>=0;--i){
c[i]=(x*base+s[i])>>1;
x=(s[i]&1);
}
c.lenth();/////////////////////////////
return c;
}
bigint multi2(){//
long long x=0;
bigint c=0;
for(register int i=0;i<tot;++i){
x+=(s[i]<<1);
c[c.tot++]=x%base;
x/=base;
}
while(x){
c[c.tot++]=x%base;
x/=base;
}
return c;
}
};
bigint fmi(bigint a,int x){
bigint ans;
ans=1;
while(x){
(x&1)?(ans=ans.multiply(a)):1;
x>>=1;
a=a.multiply(a);
}
return ans;
}
int m;
char a[10005];
int main(){
// freopen("GG.out","r",stdin);
// freopen("GGout.out","w",stdout);
scanf("%d",&m);
scanf("%s",a);
bigint GG,r,l,mid,one,ans;
one=1;
GG=a;
r=2;
l=1;
if(m==1){
GG.print();
return 0;
}
if(GG<one){
putchar('0');
return 0;
}
while(1){
if(fmi(r,m)<GG){
l=r;
r=r.multi2();
}
else break;
}
while(l<r){
mid=(l+r).divide();
if(fmi(mid,m)<=GG){
ans=mid;
l=mid+one;
}
else r=mid-one;
}
long long zzzzz=0;
bigint zzzzzz;
zzzzzz=zzzzz;
if(ans==zzzzzz){
printf("1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000");
return 0;
}
ans.print();
// fclose(stdin);
// fclose(stdout);
return 0;
}
高精，二分答案，快速幂.
极其恶(nao)心(can)的是，n>=0，这要是考试可就真GG了.
最后一个点莫名其妙WA了就搞了一大大大波特判
禁止诸如开发者知识库/布布扣/码迷/学步园/马开东等 copy 他人博文乃至博客的网站转载
，用户转载请注明出处：https://www.cnblogs.com/xcysblog/
浙公网安备 33010602011771号