快速读入输出
namespace io{
#define gc() (iS == iT ? (iT = (iS = ibuff) + fread(ibuff, 1, SIZ, stdin), (iS == iT ? EOF : *iS++)) : *iS++)
const int SIZ = 1 << 21 | 1;
char *iS, *iT, ibuff[SIZ], obuff[SIZ], *oS = obuff, *oT = oS + SIZ - 1, fu[110], c;
int fr;
inline void out() {
fwrite(obuff, 1, oS - obuff, stdout);
oS = obuff;
}
template<class Type>
inline void read(Type &x) {
x = 0;
Type y = 1;
for (c = gc(); (c > '9' || c < '0') && c != '-'; c = gc());
c == '-' ? y = -1 : x = (c & 15);
for (c = gc(); c >= '0' && c <= '9'; c = gc())
x = x * 10 + (c & 15);
x *= y;
}
template<class Type>
inline void print(Type x, char text = '\n') {
if (x < 0)
*oS++ = '-', x *= -1;
if (x == 0)
*oS++ = '0';
while (x)
fu[++fr] = x % 10 + '0', x /= 10;
while (fr)
*oS++ = fu[fr--];
*oS++ = text;
out();
}
}
using namespace io;
大整数类
namespace BigInteger
{
#define maxn 10005
using std::sprintf;
using std::string;
using std::max;
using std::istream;
using std::ostream;
struct Big_integer{
int d[maxn], len;
void clean()
{
while(len > 1 && !d[len-1])
len--;
}
Big_integer() { memset(d, 0, sizeof(d)); len = 1; }
Big_integer(int num) { *this = num; }
Big_integer(char* num) { *this = num; }
Big_integer operator = (const char* num){
memset(d, 0, sizeof(d)); len = strlen(num);
for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';
clean();
return *this;
}
Big_integer operator = (int num){
char s[10005]; sprintf(s, "%d", num);
*this = s;
return *this;
}
Big_integer operator + (const Big_integer& b){
Big_integer c = *this; int i;
for (i = 0; i < b.len; i++){
c.d[i] += b.d[i];
if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++;
}
while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;
c.len = max(len, b.len);
if (c.d[i] && c.len <= i) c.len = i+1;
return c;
}
Big_integer operator - (const Big_integer& b){
Big_integer c = *this; int i;
for (i = 0; i < b.len; i++){
c.d[i] -= b.d[i];
if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;
}
while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;
c.clean();
return c;
}
Big_integer operator * (const Big_integer& b)const{
int i, j; Big_integer c; c.len = len + b.len;
for(j = 0; j < b.len; j++) for(i = 0; i < len; i++)
c.d[i+j] += d[i] * b.d[j];
for(i = 0; i < c.len-1; i++)
c.d[i+1] += c.d[i]/10, c.d[i] %= 10;
c.clean();
return c;
}
Big_integer operator / (const Big_integer& b){
int i, j;
Big_integer c = *this, a = 0;
for (i = len - 1; i >= 0; i--)
{
a = a*10 + d[i];
for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
c.d[i] = j;
a = a - b*j;
}
c.clean();
return c;
}
Big_integer operator % (const Big_integer& b){
int i, j;
Big_integer a = 0;
for (i = len - 1; i >= 0; i--)
{
a = a*10 + d[i];
for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
a = a - b*j;
}
return a;
}
Big_integer operator += (const Big_integer& b){
*this = *this + b;
return *this;
}
bool operator <(const Big_integer& b) const{
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--)
if(d[i] != b.d[i]) return d[i] < b.d[i];
return false;
}
bool operator >(const Big_integer& b) const{return b < *this;}
bool operator<=(const Big_integer& b) const{return !(b < *this);}
bool operator>=(const Big_integer& b) const{return !(*this < b);}
bool operator!=(const Big_integer& b) const{return b < *this || *this < b;}
bool operator==(const Big_integer& b) const{return !(b < *this) && !(b > *this);}
string str() const{
char s[maxn]={};
for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';
return s;
}
};
istream& operator >> (istream& in, Big_integer& x)
{
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream& out, const Big_integer& x)
{
out << x.str();
return out;
}
}
using namespace BigInteger;
图论算法
int cnt,head[100],nxt[100],to[100],edge[100];
int vis[1000010],dis[1000010];
int inq[1000010],dis[1000010];
边表存图
inline void add(int x,int y,int x){
cnt++;
nxt[cnt]=head[x];
head[x]=cnt;
to[cnt]=y;
edge[cnt]=z;
}
for(int i=head[x];i;i=nxt[i])
SPFA
inline void SPFA(int s){
for(int i=1;i<=n;i++){
dis[i]=0x7fffffff;
}
queue<int> Q;
Q.push(s);
inq[s]=1;
dis[s]=0;
while(!Q.empty()){
int fr=Q.front();
Q.pop();
inq[fr]=0;
for(int i=head[fr];i;i=nxt[i]){
int v=to[i],w=edge[i];
if(dis[fr]+w<dis[v]){
dis[v]=dis[fr]+w;
if(!inq[v])Q.push(v),inq[v]=1;
}
}
}
}
DIJKSTRA
struct node{
int u,d;
bool operator<(const node & rhs){
return d>rhs.d;
}
}temp;
inline void DIJKSTRA(int s){
for(int i=1;i<=n;i++){
dis[i]=0x7fffffff;
}
priority_queue<node> Q;
Q.push((node){s,0});
while(!Q.empty()){
temp=Q.top();
Q.pop();
int u=temp.u;
if(vis[u]){
continue;
}
vis[u]=1;
for(int i=head[u];i;i=nxt[i]){
int w=edge[i],v=to[i];
if(dis[u]+w<dis[v]){
dis[v]=dis[u]+w;
}
Q.push((node){v,dis[v]});
}
}
}
数学&&其他
辗转相除求gcd
inline int gcd(int a,int b){
return b?gcd(b,a%b):a;
}
快速幂卡粟米
inline int pow(int a,int b){
int base=a,ans=1,temp=b;
while(temp){
if(temp&1){
ans*=base;
}
temp>>1;
base*=base;
}
return ans;
}
欧拉质数筛法
inline void prime(int n) {
v[0] = v[1] = 1;
for (int i = 2; i <= n; i++) {
if (!v[i]) {
isin[i] = 1;
v[i] = 1;
for (int j = 1; j * i <= n; j++)
v[i * j] = 1;
}
}
}
归并排序
inline void marge(int a[],int l,int mid,int r,int temp[]){
int i=l,j=mid+1,n=mid,m=r,k=l;
while(i<=n&&j<=m){
if(a[i]<=a[j]){
temp[k++]=a[i++];
}else{
temp[k++]=a[j++];
}
}
while(i<=n){
temp[k++]=a[i++];
}
while(j<=n){
temp[k++]=a[j++];
}
for(int i=l;i<=r;i++){
a[i]=temp[i];
}
}
inline void msort(int a[],int l,int r,int temp[]){
if(l<r){
int mid=(l+r)/2;
msort(a,l,mid,temp);
msort(a,mid+1,r,temp);
marge(a,l,mid,r,temp);
}
}
