HDOJ 4893 Wow! Such Sequence!

题意是这样的，给定一个n个元素的数组，初始值为0，3种操作：

1 k d将第k个数增加d；

2 l r 询问区间l...r范围内数之和；

3 l r 表示将区间l...r内的数变成离他最近的斐波那契数，要求尽量小。

线段树操作题目，其中对于第三种操作用一个懒惰标记一下，表示l...r内的数是不是已经变成斐波那契数，如果是的话，求和就是其相应数的斐波那契数之和。

代码：

//Template updates date: 20140718
#include <bits/stdc++.h>
#define  esp 1e-6
#define  inf 0x3f3f3f3f
#define  pi acos(-1.0)
#define  pb push_back
#define  lson l, m, rt<<1
#define  rson m+1, r, rt<<1|1
#define  lowbit(x) (x&(-x))
#define  mp(a, b) make_pair((a), (b))
#define  bit(k) (1<<(k))
#define  iin  freopen("pow.in", "r", stdin);
#define  oout freopen("pow.out", "w", stdout);
#define  in  freopen("solve_in.txt", "r", stdin);
#define  out freopen("solve_out.txt", "w", stdout);
#define  bug puts("********))))))");
#define  Inout iin oout
#define  inout in out

#define  SET(a, v) memset(a, (v), sizeof(a))
#define  SORT(a)   sort((a).begin(), (a).end())
#define  REV(a)    reverse((a).begin(), (a).end())
#define  READ(a, n) {REP(i, n) cin>>(a)[i];}
#define  REP(i, n) for(int i = 0; i < (n); i++)
#define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
#define  Rep(i, base, n) for(int i = (base); i < (n); i++)
#define  REPS(s, i) for(int i = 0; (s)[i]; i++)
#define  pf(x) ((x)*(x))
#define  mod(n) ((n))
#define  Log(a, b) (log((double)b)/log((double)a))
#define Srand() srand((int)time(0))
#define random(number) (rand()%number)
#define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)

using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<PII, int> VIII;
typedef VI:: iterator IT;
typedef map<string, int> Mps;
typedef map<int, int> Mpi;
typedef map<int, PII> Mpii;
typedef map<PII, int> Mpiii;
const int maxm = 140000 + 1000;

LL f[92];
int n, m;
LL sum[maxm<<2], sum1[maxm<<2], cover[maxm<<2];

void pre() {
f[0] = f[1] = 1;
Rep(i, 2, 92) {
f[i] = f[i-1] + f[i-2];
}
}
void PushDown(int rt) {
if(cover[rt]) {
cover[rt] = 0;
cover[rt<<1] = cover[rt<<1|1] = 1;
sum[rt<<1] = sum1[rt<<1];
sum[rt<<1|1] = sum1[rt<<1|1];
}
}
void PushUp(int rt) {
sum[rt] = sum[rt<<1]+sum[rt<<1|1];
sum1[rt] = sum1[rt<<1]+sum1[rt<<1|1];
}
void build(int l, int r, int rt) {
if(l == r) {
sum[rt] = 0;
sum1[rt] = 1;
cover[rt] = 0;
return ;
}
int m = (l+r)>>1;
build(lson);
build(rson);
PushUp(rt);
cover[rt] = 0;
}

void update(int l, int r, int rt, int k, int d) {
if(l == k && r == k) {
sum[rt] += d;
int b = upper_bound(f+1, f+92, sum[rt]) - f;
if(abs(f[b]-sum[rt]) >= abs(f[b-1]-sum[rt]))
sum1[rt] = f[b-1];
else
sum1[rt] = f[b];
return;
}
PushDown(rt);
int m = (l+r)>>1;
if(k <= m)
update(lson, k, d);
else update(rson, k, d);
PushUp(rt);
}
void update1(int L, int R, int l, int r, int rt) {
if(L <=l && R >= r) {
cover[rt] = 1;
sum[rt] = sum1[rt];
return;
}
PushDown(rt);
int m = (l+r)>>1;
if(L <= m)
update1(L, R, lson);
if(R > m)
update1(L, R, rson);
PushUp(rt);
}
LL query(int L, int R, int l, int r, int rt) {
if(L <= l && R >= r) {
return sum[rt];
}
int m = (l+r)>>1;
PushDown(rt);
LL ans = 0;
if(L <= m)
ans += query(L, R, lson);
if( R > m)
ans += query(L, R, rson);
return ans;
}
int main() {

pre();
while(scanf("%d%d", &n, &m) == 2) {
build(1, n, 1);
REP(i, m) {
int u, l, r;
scanf("%d%d%d", &u, &l, &r);
if(u == 2) {
printf("%I64d\n", query(l, r, 1, n, 1));
} else if(u == 1) {
update(1, n, 1, l, r);
} else {
update1(l, r, 1, n, 1);
}
}
}
return 0;
}

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