模板总结(个人向）

1|0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;

2|0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;

3|0图论算法

int cnt,head[100],nxt[100],to[100],edge[100];
int vis[1000010],dis[1000010];
int inq[1000010],dis[1000010];

3|1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])

3|2SPFA

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;
            }
        }
    }
}

3|3DIJKSTRA

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]});
        }
    }
}

4|0数学&&其他

4|1辗转相除求gcd

inline int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}

4|2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;
}

4|3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;
        }
    }
}

4|4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);
    }
}

__EOF__

本文作者：Kdlyh
本文链接：https://www.cnblogs.com/kdlyh/p/17777003.html
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