HDU 5239 Doom The 2015 ACM-ICPC China Shanghai Metropolitan Programming Contest
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题意
给定一段序列,你有一个sum(初始值为0),只有一种操作(l,r),就是将序列中从l到r区间内的数字加到sum上,之后将每个数字平方,每次操作输出一次sum的值。
题解
本来还是一道比较简单的线段树,但是有两个坑点。
1.给出的mod值非常大为2^63-2^31。当你进行平方运算的时候一定会溢出,所以乘法运算得用快速加法解决(mod+mod < 2^64)
2.由于MOD很大,会发现对于一个区间,在询问最多30次之后,会变成一个固定不变的数值。(需要打个表尝试T_T。。。)也就是对于任何一个区间,我们更新了30次之后就不再更新了。(这个必须有!!!否则TLE妥妥的)
剩下的东西就非常好理解了,贴一下我的AC代码~
#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#define INF 9223372034707292160#define ll unsigned __int64using namespace std;ll n,m,b[1000005],sum;struct node{ ll l,r,sum,add;}a[1001000<<1];void pushup(ll root){ a[root].sum = (a[root<<1].sum + a[root<<1|1].sum)%INF; a[root].add = min(a[root<<1].add,a[root<<1|1].add);}void build_tree(ll root,ll l,ll r){ a[root].l = l; a[root].r = r; if(l == r){ a[root].sum = b[l]; a[root].add = 0; return ; } ll mid = (l + r) >> 1; build_tree(root<<1,l,mid); build_tree(root<<1|1,mid+1,r); pushup(root);}ll fun(ll a,ll b){ ll sum = 0; while(b){ if(b&1){ sum = (sum + a) % INF; } b /= 2; a = (2*a) % INF; } return sum;}void update(ll root,ll x,ll y){ ll l = a[root].l; ll r = a[root].r; if(l == x && r == y && x == y){## 标题 ## a[root].sum = fun(a[root].sum,a[root].sum); a[root].add++; return ; } ll mid = (l + r) >> 1; if(x > mid) update(root<<1|1,x,y); else if(y <= mid) update(root<<1,x,y); else{ update(root<<1|1,mid+1,y); update(root<<1,x,mid); } pushup(root);}ll query(ll root,ll x,ll y){ ll l = a[root].l; ll r = a[root].r; if(l == x && r == y){ return a[root].sum; } ll mid = (l + r) >> 1; if(y <= mid) return (query(root << 1,x,y))%INF; else if(x > mid) return (query(root << 1|1,x,y))%INF; else{ return (query(root << 1,x,mid)+query(root<<1|1,mid+1,y))%INF; }}int main(){ int i,j,T,xxx = 1; ll x,y; cin >> T; while(T--){ scanf("%I64d%I64d",&n,&m); for(i=1;i<=n;i++){ scanf("%I64d",&b[i]); } build_tree(1,1,n); printf("Case #%d:\n",xxx++); sum = 0; for(i=0;i<m;i++){ scanf("%I64d%I64d",&x,&y); sum = (sum+query(1,x,y))%INF; printf("%I64d\n",sum); update(1,x,y); } } return 0;}
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