CSU 1720 How to Get 2^n （大数+hash）

题意：给你10W个数字，每个数都是大数，范围是1到10^30，然后问你有多少种方法，每次选取两个数，两个数的和是2的幂次

题解：10的30次大约是2的100次，所以先预处理2的102次，然后就是每次输入一个大数，枚举2的幂次去减它，然后去map里找有多少个解，其实是个很简单的思路，但是我却一直写炸，主要是大数的模板太差，会T，加上我智商下线，开了很大的数组去存输入的内容，结果实力T。

其实一边输入，然后转化为大数，然后枚举2的幂次，相减，然后hash，去map里面查，然后累计求和，然后把输入hash，放进map里，这样就好了

问qwb神犇要了新的姿势的大数模板otz，感觉十分之屌，跑得飞快

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define MAX 100005
//#define MAXN 500005
#define maxnode 105
#define sigma_size 2
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000

//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const ull mx = 1e9+7;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

const int MX = 10;//一共可以表示长度MX*DLEN的
const int maxn = 9999;
const int DLEN = 4;//一个int里面放多少个数字

char ret[105];
class Big {
public:
int a[MX], len;
Big(const int b = 0) {
int c, d = b;
len = 0;
memset(a, 0, sizeof(a));
while(d > maxn) {
c = d - (d / (maxn + 1)) * (maxn + 1);
d = d / (maxn + 1);
a[len++] = c;
}
a[len++] = d;
}
Big(const char *s) {
int t, k, index, L, i;
memset(a, 0, sizeof(a));
L = strlen(s);
len = L / DLEN;
if(L % DLEN) len++;
index = 0;
for(i = L - 1; i >= 0; i -= DLEN) {
t = 0;
k = i - DLEN + 1;
if(k < 0) k = 0;
for(int j = k; j <= i; j++) {
t = t * 10 + s[j] - '0';
}
a[index++] = t;
}
}
Big operator/(const int &b)const {
Big ret;
int i, down = 0;
for(int i = len - 1; i >= 0; i--) {
ret.a[i] = (a[i] + down * (maxn + 1)) / b;
down = a[i] + down * (maxn + 1) - ret.a[i] * b;
}
ret.len = len;
while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
return ret;
}
bool operator>(const Big &T)const {
int ln;
if(len > T.len) return true;
else if(len == T.len) {
ln = len - 1;
while(a[ln] == T.a[ln] && ln >= 0) ln--;
if(ln >= 0 && a[ln] > T.a[ln]) return true;
else return false;
} else return false;
}
Big operator+(const Big &T)const {
Big t(*this);
int i, big;
big = T.len > len ? T.len : len;
for(i = 0; i < big; i++) {
t.a[i] += T.a[i];
if(t.a[i] > maxn) {
t.a[i + 1]++;
t.a[i] -= maxn + 1;
}
}
if(t.a[big] != 0) t.len = big + 1;
else t.len = big;
return t;
}
Big operator-(const Big &T)const {
int i, j, big;
bool flag;
Big t1, t2;
if(*this > T) {
t1 = *this;
t2 = T;
flag = 0;
} else {
t1 = T;
t2 = *this;
flag = 1;
}
big = t1.len;
for(i = 0; i < big; i++) {
if(t1.a[i] < t2.a[i]) {
j = i + 1;
while(t1.a[j] == 0) j++;
t1.a[j--]--;
while(j > i) t1.a[j--] += maxn;
t1.a[i] += maxn + 1 - t2.a[i];
} else t1.a[i] -= t2.a[i];
}
t1.len = big;
while(t1.a[t1.len - 1] == 0 && t1.len > 1) {
t1.len--;
big--;
}
if(flag) t1.a[big - 1] = 0 - t1.a[big - 1];
return t1;
}
int operator%(const int &b)const {
int i, d = 0;
for(int i = len - 1; i >= 0; i--) {
d = ((d * (maxn + 1)) % b + a[i]) % b;
}
return d;
}
Big operator*(const Big &T) const {
Big ret;
int i, j, up, temp, temp1;
for(i = 0; i < len; i++) {
up = 0;
for(j = 0; j < T.len; j++) {
temp = a[i] * T.a[j] + ret.a[i + j] + up;
if(temp > maxn) {
temp1 = temp - temp / (maxn + 1) * (maxn + 1);
up = temp / (maxn + 1);
ret.a[i + j] = temp1;
} else {
up = 0;
ret.a[i + j] = temp;
}
}
if(up != 0) {
ret.a[i + j] = up;
}
}
ret.len = i + j;
while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
return ret;
}
ull print() {
ull ret=0;
for(int i=len-1;i>=0;i--) ret=ret*mx+a[i];
return ret;
}
}pow2[105],ss;
map<ull,int> ma;
char s[105];
void init(){
pow2[0]=1;
for(int i=1;i<=103;i++) pow2[i]=pow2[i-1]*2;
}
int main(){
//freopen("test.txt","r",stdin);
int t;
cin>>t;
init();
while(t--){
int n;
cin>>n;
ma.clear();
LL ans=0;
for(int i=0;i<n;i++){
scanf("%s",s);
int len;
ss=s;
for(int j=103;j>=1;j--){
if(pow2[j]>ss){
Big cnt=pow2[j]-ss;
ull temp=cnt.print();
if(ma.count(temp)) ans+=ma[temp];
}
}
ull tmp=ss.print();
if(!ma.count(tmp)) ma[tmp]=0;
ma[tmp]++;
//cout<<ma[ss[i]]<<" "<<ss[i]<<endl;
}
cout<<ans<<endl;
}
return 0;
}

/**************************************************************
Problem: 1720
User: 1403mashaonan
Language: C++
Result: Accepted
Time:5640 ms
Memory:4272 kb
****************************************************************/