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高精度压位模板

2022-06-21 16:03:00 eliforsharon

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#define inf 1000000000]
#define p 9 //注意乘法压位过大
using namespace std;
void _scanf(int &x){
	char ch=getchar();
	bool f=0;
	x=0;
	while(!isdigit(ch)) {
	if(ch=='-') f=1;
	ch=getchar();
	}
	while(isdigit(ch)) x=10*x+ch-'0',ch=getchar();
	if(f) x=-x;
}
void _scanf(int &a,int &b){
	_scanf(a);
	_scanf(b);
}


struct bign{
	int z[25],len;
	bign(){
		memset(z,0,sizeof(z));
		len=0;
	}
	bign (int x){
		*this=x;
	}
	bign operator =(int x){
		char s[25];
		sprintf(s,"%d",x);
		*this=s;
	}
	bign operator =(char *s){
		int len=strlen(s);
		int cur=0;
		bign a;
		char w[25];
		while(len>0){
			while(len>=p){
				strcpy(w,s+len-p);
				a.z[cur++]=atoi(w);
				len-=p;
			}
			if(len>0){
				strcpy(w,s);
				w[len]='\0';
				a.z[cur++]=atoi(w);
				len=0;
			}
		}
		a.len=cur;
		*this=a;
	}
	friend bign operator +(bign a,bign b){
		bign c;
		int len=max(a.len,b.len)+1;
		for(int i=0;i<=len-1;i++){
			c.z[i]=a.z[i]+b.z[i];
		}
		for(int i=0;i<=len-1;i++)
			c.z[i+1]+=c.z[i]/1000000000,c.z[i]%=1000000000;
		while(c.z[len-1]==0&&len>1) len--;
		c.len=len;
		return c;
	}
};

void bignout(bign a){
	int len=a.len;
	printf("%d",a.z[len-1]);
	for(int i=len-2;i>=0;i--)
		printf("%0*d",p,a.z[i]);
}

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本文为[eliforsharon]所创,转载请带上原文链接,感谢
https://blog.csdn.net/eliforsharon/article/details/77725529