高精度压位

高精度

 转载于：https://blog.youkuaiyun.com/devillaw_zhc/article/details/7776578

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define re register int
using namespace std;
const int power = 4;
const int base = 1e4;
const int L = 1001;
char str1[L],str2[L];
struct num{
	int a[L];
	num(){memset(a,0,sizeof(a));}
	num(char *s){
		memset(a,0,sizeof(a));
		int len = strlen(s);
		a[0] = (len+power-1)/power;
		for(re i=0,t=0,w;i<len;w*=10,i++){
			if(i%power == 0){w = 1; t++;}
			a[t]+=w*(s[i]-'0');
		}
	}
	void add(int k){if(k||a[0]){a[++a[0]] = k;}}//除法时用
	void rever(){reverse(a+1,a+a[0]+1);} //除法时用 
	void print(){
		printf("%d",a[ a[0] ]);
		for(re i=a[0]-1;i;i--){
			printf("%0*d",power,a[i]);// %0*d 必须输出power位（补零） 
		}
		printf("\n");
	} 
}p,q,ans;
bool operator <(const num &p,const num &q){
	if(p.a[0]<q.a[0]) return true;
	else if(p.a[0]>q.a[0]) return false;
	for(re i=p.a[0];i;i--){
		if(p.a[i]!=q.a[i]) return p.a[i]<q.a[i];
	}
	return false;
}
num operator +(const num &p,const num &q){
	num c;
	c.a[0] = max(p.a[0],q.a[0]);
	for(re i=1; i<=c.a[0];i++){
		c.a[i] += p.a[i]+q.a[i];
		c.a[i+1] += c.a[i]/base;
		c.a[i] %=base;
	}
	if(c.a[c.a[0]+1])++c.a[0];
	return c;
}
num operator -(const num &p,const num &q){
	num c = p;
	for(re i=1; i<=c.a[0]; i++){
		c.a[i] -=q.a[i];
		if(c.a[i]<0){
			c.a[i]+=base;
			c.a[i+1]--;
		}
	}
	while(c.a[0]>0&&!c.a[c.a[0]]) --c.a[0];
	return c;
}
num operator *(const num &p,const num &q){
	num c;
	c.a[0] = p.a[0]+q.a[0]-1;
	for(re i=1; i<=p.a[0];i++){
		for(re j=1; j<=q.a[0]; j++){
			c.a[i+j-1] += p.a[i]*q.a[j];
			c.a[i+j] += c.a[i+j-1]/base;
			c.a[i+j-1] %= base;
		}
	}
	if(c.a[c.a[0]+1]) ++c.a[0];
	return c;
}
num operator /(const num &p, const num &q){
	num x,y;
	for(re i=p.a[0];i>=1;i--){
		y.add(p.a[i]);
		y.rever();
		while(!(y<q)){
			y = y-q;
			++x.a[i];
		}
		y.rever();
	}
	x.a[0] = p.a[0];
	while(x.a[0]>0&&!x.a[x.a[0]]) --x.a[0];
	return x;
}
int main(){
	scanf("%s",str1);
	scanf("%s",str2);
	reverse(str1,str1+strlen(str1));
	reverse(str2,str2+strlen(str2));
	p = num(str1);
	q = num(str2);
	ans = p+q;
	ans.print();
	ans = p-q;
	ans.print();
	ans = p*q;
	ans.print();
	ans = p/q;
	ans.print();
} 

洛谷P1005 矩阵取数游戏（真心觉得坑，2^80比unsigned long long大一点，用个高精真心烦。）

模板针对这题增加了两个函数

reset()重置数组

num_er()初始化数组

	void reset(){memset(a,0,sizeof(a));}
	void num_er(int k){
		a[0] = 1;
		a[1] = 1;
		return ;
	}

int n,m;
num a[L][L],er[81],f[81][81],ta,tb,ans;
char ch[L];
int main(){
	scanf("%d%d",&n,&m);
	for(re i=1; i<=n; i++){
		for(re j=1; j<=m; j++){
			scanf("%s",ch);
			reverse(ch,ch+strlen(ch));
			a[i][j] = (ch);
		}
	}
	er[0].num_er(1);
	for(re i=1; i<=m; i++){
		er[i] = er[i-1]+er[i-1];
	}
	for(re k=1; k<=n; k++){
		for(re i=1; i<=m; i++){
			for(re j=1; j<=m; j++){
				f[i][j].reset();
			}
		}
		for(re i=1; i<=m; i++){
			f[i][i] = f[i][i]+a[k][i];
			f[i][i] = f[i][i]*er[m];
		}
		for(re l=1; l<m; l++){
			for(re i=1,j=i+l; j<=m; i++,j=i+l){
				ta.reset(); tb.reset(); 
				ta = ta+a[k][i];
				ta = ta*er[m+i-j];
				ta = ta+f[i+1][j];
				
				tb = tb+a[k][j];
				tb = tb*er[m+i-j];
				tb = tb+f[i][j-1];
				if(tb<ta){
					f[i][j] = ta;
				}else{
					f[i][j] = tb;
				}
			}
		}
		ans = ans+f[1][m];
	}
	ans.print();
	return 0;
}