HDU 4893 Wow! Such Sequence! (树状数组)

题意：给有三种操作，一种是 1 k d，把第 k 个数加d，第二种是2 l r，查询区间 l, r的和，第三种是 3 l r，把区间 l，r 的所有数都变成离它最近的Fib数，

并且是最小的那个。

析：觉得应该是线段树的，但是。。。不会啊。。。就想胡搞一下。

所以用了树状数组，也就是和的，然后用一个set来维护每个不是Fibnoccia的数，然后再进行计算。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2};
const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL sum1[maxn<<1], a[maxn];
LL F[150];
int cnt;
set<int> sets;
set<int> :: iterator it;
int lowbit(int x){ return x & (-x); }
 
void add1(int x, LL d){
    while(x <= n){
        sum1[x] += d;
        x += lowbit(x);
    }
}
 
LL qurey1(int x){
    LL ans = 0;
    while(x > 0){
        ans += sum1[x];
        x -= lowbit(x);
    }
    return ans;
}
 
void solve(int l, int r){
    it = sets.lower_bound(l);
    while(it != sets.end() && *it <= r){
        LL *tmp = lower_bound(F+1, F+cnt, a[*it]);
        if(a[*it] == *tmp){ sets.erase(it++); continue; }
        LL *tmpp = tmp - 1;
        if(*tmp - a[*it] >= a[*it] - *tmpp){
            add1(*it, *tmpp - a[*it]);
            a[*it] = *tmpp;
            sets.erase(it++);
        }
        else{
            add1(*it, *tmp - a[*it]);
            a[*it] = *tmp;
            sets.erase(it++);
        }
    }
}
int main(){
    F[0] = F[1] = 1;
    cnt = 2;
    while(1){
        F[cnt] = F[cnt-1] + F[cnt-2];
        if(F[cnt] > (1LL<<61))  break;
        ++cnt;
    }
    ++cnt;
    while(scanf("%d %d", &n, &m) == 2){
        sets.clear();
        for(int i = 0; i <= n; ++i){
            sum1[i] = a[i] = 0;
            sets.insert(i);
        }
        int l, r, x;
        for(int i = 0; i < m; ++i){
            scanf("%d", &x);
            if(1 == x){
                scanf("%d %d", &l, &r);
                a[l] += r;
                sets.insert(l);
                add1(l, (LL)r);
            }
            else if(2 == x){
                scanf("%d %d", &l, &r);
                printf("%I64d\n", qurey1(r) - qurey1(l-1));
            }
            else if(3 == x){
                scanf("%d %d", &l, &r);
                solve(l, r);
            }
        }
    }
    return 0;
}