hdu 5239 区间平方 线段树区间更新

http://acm.hdu.edu.cn/showproblem.php?pid=5239

题意：给出n个数和一个初始值为0的答案。每次操作给出一个区间[l,r]，把区间所有的数加到答案中，之后把区间的每个数都平方。每次操作都需要输出答案 mod 9223372034707292160(2 ^ 63 - 2 ^ 31)

对于p这个数，任何一个数平方对p这个数取模最多29次，之后就会是不变的数，所以更新时判断更新次数是否等于30，如果是则不用向下更新，否则继续向下更新。

需要注意的是，当(p-1)*(p-1)时，直接相乘会溢出，所以此处要用到快速加法

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll unsigned __int64
#define mp make_pair
#define pb push_back
typedef ll LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
//const int dx[]={-1,0,1,0,1,-1,-1,1};
//const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=10086;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(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;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}

const LL p=9223372034707292160uLL;
LL a[maxn];
int n,m;
LL ans=0;
struct node
{
LL l,r,sum,lazy;
}tree[40*maxn];
void pushup(int id)
{
tree[id].sum=(tree[id<<1].sum+tree[id<<1|1].sum)%p;
tree[id].lazy=min(tree[id<<1].lazy,tree[id<<1|1].lazy);
}
void build(int id,int l,int r)
{
tree[id].l=l;
tree[id].r=r;
if(l==r)
{
tree[id].sum=a[l];
tree[id].lazy=0;
return ;
}
LL mid=(l+r)>>1;
build(id<<1,l,mid);
build(id<<1|1,mid+1,r);
pushup(id);
}

LL fun(LL a,LL b)
{
LL sum=0;
W(b)
{
if(b&1)
sum=(sum+a)%p;
b/=2;
a=a*2%p;
}
return sum;
}

void update(int id,int l,int r)
{
if(tree[id].lazy==30)
return ;
if(l==tree[id].l&&r==tree[id].r&&l==r)
{
tree[id].sum=fun(tree[id].sum,tree[id].sum);
tree[id].lazy++;
return ;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(l>mid)
update(id<<1|1,l,r);
else if(r<=mid)
update(id<<1,l,r);
else
{
update(id<<1|1,mid+1,r);
update(id<<1,l,mid);
}
pushup(id);
}

LL query(int id,int l,int r)
{
if(l<=tree[id].l&&r>=tree[id].r)
return tree[id].sum;
int mid=(tree[id].l+tree[id].r)>>1;
LL ans=0;
if(l<=mid)
{
ans+=query(id<<1,l,r)%p;
ans%=p;
}
if(r>mid)
{
ans+=query(id<<1|1,l,r)%p;
ans%=p;
}
return ans;
}

void solve()
{
s_2(n,m);
FOR(1,n,i) S_1(a[i]);
build(1,1,n);
ans=0;
W(m--)
{
int l,r;
s_2(l,r);
ans=(ans+query(1,l,r)%p)%p;
print(ans);
update(1,l,r);
}
}
int main()
{
//freopen( "1.txt" , "r" , stdin );
//freopen( "2.in" , "w" , stdout );
int t=1;
//init();
s_1(t);
for(int cas=1;cas<=t;cas++)
{
printf("Case #%d:\n",cas);
solve();
}
}