标签：des style blog http color java os strong
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2234 Accepted Submission(s): 657
【题目分析】
这题相比于裸的线段树区间更新有了一些难度。
我们在每个结点中设一个fib，表示离sum最近的fibnacci数，每次区间更新时，就将sum的值更新为fib。fib的值只有在单点更新的过程中才会改变，也就是说当sum值改变的时候fib才改变，因为当sum变为fib后，离sum最近的fibnacci数还是fib值。
lazy----记录该点以下的孩子结点是否需要更新。
我用long long ,然后用了%lld输入，不知道杭电不支持%lld，debug了半天，TLE了20多次，后来改%lld为%I64d就过了，哭死QAQ...
//Memory Time
// K MS
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 100010
#define LL long long
using namespace std;
LL n,m;
LL ans;
LL f[60];
struct Tree
{
LL l,r;
LL sum,fib;
bool lazy;
};
Tree tree[MAX*3];
LL Find(LL x)
{
if(x<=0)return 1;
LL ldis,rdis;
for(int i=0;i<59;++i)
{
if(f[i]<=x&&f[i+1]>=x)
{
ldis=x-f[i];
rdis=f[i+1]-x;
return ldis<=rdis?f[i]:f[i+1];
}
}
}
void pushup(LL x)
{
LL tmp=x<<1;
tree[x].sum=tree[tmp].sum+tree[tmp+1].sum;
tree[x].fib=tree[tmp].fib+tree[tmp+1].fib;
}
void pushdown(LL x)
{
if(!tree[x].lazy)return;
tree[x].lazy=0;
if(tree[x].l==tree[x].r)return;
LL tmp=x<<1;
tree[tmp].lazy=tree[tmp+1].lazy=1;
tree[tmp].sum=tree[tmp].fib;
tree[tmp+1].sum=tree[tmp+1].fib;
}
void build(LL l,LL r,LL x)
{
tree[x].l=l,tree[x].r=r;
tree[x].sum=0,tree[x].fib=1,tree[x].lazy=0;
if(l==r)return;
LL tmp=x<<1;
LL mid=(l+r)>>1;
build(l,mid,tmp);
build(mid+1,r,tmp+1);
pushup(x);
}
void add(LL x,LL k,LL num)
{
if(tree[x].l==tree[x].r)
{
tree[x].sum+=num;
tree[x].fib=Find(tree[x].sum);
return;
}
if(tree[x].lazy)
pushdown(x);
LL tmp=x<<1;
LL mid=(tree[x].l+tree[x].r)>>1;
if(k<=mid)
add(tmp,k,num);
else if(k>mid)
add(tmp+1,k,num);
pushup(x);
}
void change(LL l,LL r,LL x)
{
if(r<tree[x].l||l>tree[x].r)return;
if(l<=tree[x].l&&r>=tree[x].r)
{
tree[x].sum=tree[x].fib;
tree[x].lazy=1;
return;
}
if(tree[x].lazy)pushdown(x);
LL tmp=x<<1;
LL mid=(tree[x].l+tree[x].r)>>1;
if(r<=mid)
change(l,r,tmp);
else if(l>mid)
change(l,r,tmp+1);
else
{
change(l,mid,tmp);
change(mid+1,r,tmp+1);
}
pushup(x);
}
void query(LL l,LL r,LL x)
{
if(r<tree[x].l||l>tree[x].r)return;
if(l<=tree[x].l&&r>=tree[x].r)
{
ans+=tree[x].sum;
return;
}
if(tree[x].lazy)
pushdown(x);
LL tmp=x<<1;
LL mid=(tree[x].l+tree[x].r)>>1;
if(r<=mid)
query(l,r,tmp);
else if(l>mid)
query(l,r,tmp+1);
else
{
query(l,mid,tmp);
query(mid+1,r,tmp+1);
}
pushup(x);
}
int main()
{
// freopen("cin.txt","r",stdin);
// freopen("cout.txt","w",stdout);
f[0]=f[1]=1;
for(int i=2;i<60;++i) f[i]=f[i-1]+f[i-2];
while(scanf("%I64d %I64d",&n,&m)!=EOF)
{
build(1,n,1);
LL a,b,c;
while(m--)
{
scanf("%I64d %I64d %I64d",&a,&b,&c);
if(a==1)
add(1,b,c);
else if(a==2)
{
ans=0;
query(b,c,1);
printf("%I64d\n",ans);
}
else
change(b,c,1);
}
}
return 0;
}
线段树 + 区间更新： HDU 4893 Wow! Such Sequence!,布布扣,bubuko.com
线段树 + 区间更新： HDU 4893 Wow! Such Sequence!
标签：des style blog http color java os strong
原文地址：http://www.cnblogs.com/acmer-jsb/p/3884474.html
