HDU 5245 Joyful(線段樹、查詢區間和及修改區間每個數為數的平方)
阿新 • • 發佈:2018-12-24
題目連結:
HDU 5245 Joyful
題意:
查詢區間和及修改查詢區間,將每個數都變為數的平方。模數是
分析:
任何一個數的若干次平方後模上
用
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <limits.h>
#include <time.h>
#include <string>
#define lson(x) (x<<1)
#define rson(x) ((x<<1)|1)
//#define ll unsigned long long
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef unsigned long long ll;
const ll mod =9223372034707292160uLL; //需要用unsigned long long 儲存模數
const int maxn=100500;
int T,n,m,cases=0;
ll ans,val[maxn];
struct SegTree{
int left,right,flag;
ll val;
}segtree[maxn<<2];
//快速乘法
inline ll mul(ll x,ll y)
{
ll res=0;
while(y){
if(y&1) res=(res+x)%mod;
y>>=1;
x=x*2%mod;
}
return res;
}
inline void build(int left,int right,int cur)
{
segtree[cur].left=left;
segtree[cur].right=right;
segtree[cur].flag=0;
if(left==right)
{
segtree[cur].val=val[left];
return;
}
int mid=(left+right)>>1;
build(left,mid,(cur<<1));
build(mid+1,right,(cur<<1)|1);
segtree[cur].val=segtree[(cur<<1)].val+segtree[((cur<<1)|1)].val;
}
inline void query(int a,int b,int cur)
{
int left=segtree[cur].left;
int right=segtree[cur].right;
if(a==left && b==right && segtree[cur].flag==1) { //判斷當前區間是否是查詢區間並且是否已標記
ans=(ans+segtree[cur].val)%mod;
return ;
}
if(left==right)
{
ans=(ans+segtree[cur].val)%mod;
ll tmp=mul(segtree[cur].val,segtree[cur].val);
if(tmp==segtree[cur].val){ //平方後資料不變
segtree[cur].flag=1; //標記flag為1
}
segtree[cur].val=tmp;
return ;
}
int mid=(left+right)>>1;
if(b<=mid) query(a,b,(cur<<1));
else if(a>mid) query(a,b,((cur<<1)|1));
else {
query(a,mid,(cur<<1));
query(mid+1,b,((cur<<1)|1));
}
segtree[cur].val=(segtree[(cur<<1)].val+segtree[((cur<<1)|1)].val)%mod;
//如果左右子區間的資料平方後都不變,標記當前區間
if(segtree[cur<<1].flag && segtree[(cur<<1)|1].flag ) segtree[cur].flag=1;
}
int main()
{
freopen("D.in","r",stdin);
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) {
scanf("%I64u",&val[i]);
}
build(1,n,1);
ans=0;
printf("Case #%d:\n",++cases);
for(int i=0;i<m;i++) {
int a,b;
scanf("%d%d",&a,&b);
query(a,b,1);
ans%=mod;
printf("%I64u\n",ans);
}
}
return 0;
}