[java线段树]2015上海邀请赛 D Doom

题意：n个数 m个询问

        每个询问[l, r]的和， 再把[l, r]之间所有的数变为平方(模为9223372034707292160LL)

很明显的线段树

看到这个模(LLONG_MAX为9223372036854775807) 很明显平方时会爆LL

很容易发现所有数平方模了几次之后值就不再改变了 而且这个“几次”相当小 因此直接暴力搞就好了

    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        BigInteger a=BigInteger.valueOf(2);     //   这里看的是2的平方
        Long b=9223372034707292160L;
        BigInteger mod=new BigInteger(b.toString());
        for(int i=1;i<=40;i++)
        {
            a=a.multiply(a).mod(mod);
            System.out.println(i + ":" + a);     //  平方i次之后的值
        }
    }

import java.io.*;
import java.util.*;
import java.math.*;
import java.nio.charset.StandardCharsets;

public class Main
{
    static BigInteger li=BigInteger.ZERO;
    static Long b=9223372034707292160L;
    static BigInteger mod=new BigInteger(b.toString());
    static BigInteger[] sum=new BigInteger[400005];
    static boolean[] num=new boolean[400005];
    static InputReader in = new InputReader();
    public static void pushup(int rt)
    {
        sum[rt]=(sum[rt*2].add(sum[rt*2+1])).mod(mod);
        num[rt]=num[rt*2]&num[rt*2+1];
    }
    public static void build(int l, int r, int rt)
    {
        if(l==r)
        {
            sum[rt]=in.nextBigInteger();
            num[rt]=false;
            return ;
        }
        int m=(l+r)/2;
        build(l, m, rt*2);
        build(m+1, r, rt*2+1);
        pushup(rt);
    }
    public static BigInteger query(int L, int R, int l, int r, int rt)
    {
        if(L<=l && r<=R)
            return sum[rt];
        int m=(l+r)/2;
        BigInteger ret=li;
        if(L<=m)
            ret=ret.add(query(L, R, l, m, rt*2)).mod(mod);
        if(R>m)
            ret=ret.add(query(L, R, m+1, r, rt*2+1)).mod(mod);
        return ret.mod(mod);
    }
    public static void update(int L, int R, int l, int r, int rt)
    {
        if(num[rt])
            return ;
        if(l==r)
        {
            BigInteger cur=(sum[rt].multiply(sum[rt])).mod(mod);
            if(sum[rt].equals(cur))
                num[rt]=true;
            sum[rt]=cur;
            return ;
        }
        int m=(l+r)/2;
        if(L<=m)
            update(L, R, l, m, rt*2);
        if(R>m)
            update(L, R, m+1, r, rt*2+1);
        pushup(rt);
    }
    public static void main(String[] args)
    {
        PrintWriter out = new PrintWriter(System.out);
        int t, ca=1;
        t=in.nextInt();
        while((t--)!=0)
        {
            int n=in.nextInt();
            int m=in.nextInt();
            build(1, n, 1);
            System.out.println("Case #" + ca + ":");
            ca++;
            BigInteger ans=li;
            while((m--)!=0)
            {
                int l, r;
                l=in.nextInt();
                r=in.nextInt();
                ans=ans.add(query(l, r, 1, n, 1)).mod(mod);
                System.out.println(ans);
                update(l, r, 1, n, 1);
            }
        }
    }
}

class InputReader
{
    BufferedReader buf;
    StringTokenizer tok;
    InputReader()
    {
        buf = new BufferedReader(new InputStreamReader(System.in));
    }
    boolean hasNext()
    {
        while(tok == null || !tok.hasMoreElements()) 
        {
            try
            {
                tok = new StringTokenizer(buf.readLine());
            } 
            catch(Exception e) 
            {
                return false;
            }
        }
        return true;
    }
    String next()
    {
        if(hasNext()) 
            return tok.nextToken();
        return null;
    }
    int nextInt()
    {
        return Integer.parseInt(next());
    }
    long nextLong()
    {
        return Long.parseLong(next());
    }
    double nextDouble()
    {
        return Double.parseDouble(next());
    }
    BigInteger nextBigInteger()
    {
        return new BigInteger(next());
    }
    BigDecimal nextBigDecimal()
    {
        return new BigDecimal(next());
    }
}

C++11 有个神奇的东西叫做__int128   128位的整型，这题够了~

P.s. 这题很诡异的在HDOJ  java 过不了。。。MLE

#include <bits/stdc++.h>
using namespace std;
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1

typedef __int128 LL;
const LL mod=9223372034707292160;
const int N=1e5+5;
LL sum[N<<3];
bool num[N<<3];
template <class T>
bool read(T &ret){char c;int sgn;if(c=getchar(),c==EOF)return 0;while(c!='-' && (c<'0' || c>'9')) c = getchar();sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');ret *= sgn;return 1;}
template <class T>
void out(T x){if(x<0){putchar('-');x=-x;}if(x>9)out(x/10);putchar(x%10+'0');}
void pushup(int rt)
{
    sum[rt]=(sum[rt<<1]+sum[rt<<1|1])%mod;
    num[rt]=num[rt<<1]&num[rt<<1|1];
}
void build(int l, int r, int rt)
{
    if(l==r)
    {
//        scanf("%I64d", &sum[rt]);
        read(sum[rt]);
        num[rt]=0;
        return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
LL query(int L, int R, int l, int r, int rt)
{
    if(L<=l && r<=R)
        return sum[rt];
    int m=(l+r)>>1;
    LL ret=0;
    if(L<=m)
        ret=(ret+query(L, R, lson))%mod;
    if(R>m)
        ret=(ret+query(L, R, rson))%mod;
    return ret%mod;
}
void update(int L, int R, int l, int r, int rt)
{
    if(num[rt])
        return ;
    if(l==r)
    {
        LL cur=(sum[rt]*sum[rt])%mod;
        if(sum[rt]==cur)
            num[rt]=1;
        sum[rt]=cur;
        return ;
    }
    int m=(l+r)>>1;
    if(L<=m)
        update(L, R, lson);
    if(R>m)
        update(L, R, rson);
    pushup(rt);
}
int main()
{
    int t, ca=1;
    scanf("%d", &t);
    while(t--)
    {
        int n, m;
        scanf("%d%d", &n, &m);
        build(1, n, 1);
        printf("Case #%d:
", ca++);
        LL ans=0;
        while(m--)
        {
            int l, r;
            scanf("%d%d", &l, &r);
            ans=(ans+query(l, r, 1, n, 1))%mod;
//            printf("%I64d
", ans);
            out(ans);
            puts("");
            update(l, r, 1, n, 1);
        }
    }
    return 0;
}