HDU 4867 Xor

Given n non-negative integers , We define a statistical function as follows: 

It will be noted that the nature of F(k) is calculated with the number of choices, such that: 
1)0≤b _i≤a _i，0≤i≤n-1，b _i is a integer. 
2)b ₀ xor b ₁ xor ... xor b _n-1 = k，xor means exclusive-or operation.   
Now given A and m operations, there are two different operations: 
1)C x y: set the value of ax to y； 
 2)Q x: calculate F(x) mod P, where P = 1000000007.

Input

The first line has a number T (T ≤ 10), indicating the number of test cases. 
For each test case, the first line contains tow integers n, m, (1≤n, m≤20000), denote the n, m that appear in the above description. Then next line contains n non-negative integers denote 
(0≤a _i≤1000). 
Then next m lines. Each line is one of the follow two: 
1)C x y: set the value of a _x to y；(0≤x≤n-1，0≤y≤1000) 
2)Q x: calculate F(x) mod P, where P = 1000000007. 
 The number of the first operation is not more than 5000.

Output

For each Q operation, output the value of F(x) mod P.

Sample Input

12 53 2Q 3C 0 2Q 3Q 0C 0 3

Sample Output

32
3
修改和询问的操作，显然可以联想到线段树，问题是如何处理复杂的数据。
考虑到结果肯定很大，并且ai很小，用状态压缩的思想，先把ai转成2进制，
像数位dp一样，把ai差分成不同的长度的前缀01串，以可能出现的最大数字1023
为例，可以被拆成：
1111111111
1111111110
111111110X
11111110XX
1111110XXX
...
0XXXXXXXXX
X代表了01任选，显然，每个数字被拆分成的状态数等于其二进制1的个数。
于是把状态和数量插入线段树中，合并的时候采用暴力的双循环，再排个序判断重复即可。
至于复杂度，根据题解的说法，可以证明最后的状态数不会太多，并且修改操作小于5000，所以没问题。
#include<map>#include<ctime>#include<cmath>    #include<queue> #include<string>#include<vector>#include<cstdio>    #include<cstring>  #include<iostream>#include<algorithm>    #include<functional>using namespace std;#define ms(x,y) memset(x,y,sizeof(x))    #define rep(i,j,k) for(int i=j;i<=k;i++)    #define per(i,j,k) for(int i=j;i>=k;i--)    #define loop(i,j,k) for (int i=j;i!=-1;i=k[i])    #define inone(x) scanf("%d",&x)    #define intwo(x,y) scanf("%d%d",&x,&y)    #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)  #define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p) #define lson x<<1,l,mid#define rson x<<1|1,mid+1,r#define mp(i,j) make_pair(i,j)#define ft first#define sd secondtypedef long long LL;typedef pair<int, int> pii;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 1e5 + 10;const int M = 5e6;const double eps = 1e-10;int T, n, m, a[N], x, y;vector<pair<pii, int>> f[N], g;char s[N];void get(int x, int y){f[x].clear();f[x].push_back(mp(mp(y, 10), 1));for (int i = y; i; i -= low(i)){f[x].push_back(mp(mp(i - low(i), 10 - log(low(i)) / log(2)), low(i)));}}bool cmp(pair<pii, int>a, pair<pii, int>b){return a.ft < b.ft;}void merge(int x, int l, int r){f[x].clear(); g.clear();for (auto i : f[l]) for (auto j : f[r]){int k = i.ft.ft^j.ft.ft, L = min(i.ft.sd, j.ft.sd);k = (k >> (10 - L)) << (10 - L);g.push_back(mp(mp(k, L), 1LL * i.sd*j.sd%mod));}sort(g.begin(), g.end(), cmp);for (int i = 0, j; i < g.size(); i = j){pair<pii, int> q = mp(g[i].ft, 0);for (j = i; j < g.size() && g[i].ft == g[j].ft; j++){(q.sd += g[j].sd) %= mod;}f[x].push_back(q);}}void build(int x, int l, int r){if (l == r) { get(x, a[r]); return; }int mid = l + r >> 1;build(lson); build(rson);merge(x, x << 1, x << 1 | 1);}void change(int x, int l, int r, int u){if (l == r) { get(x, a[r]); return; }int mid = l + r >> 1;if (u <= mid) change(lson, u); else change(rson, u);merge(x, x << 1, x << 1 | 1);}int inv(int x) { return x == 1 ? 1 : 1LL * inv(mod%x)*(mod - mod / x) % mod; }int main(){for (inone(T); T; T--){intwo(n, m);rep(i, 1, n) inone(a[i]);build(1, 1, n);while (m--){scanf("%s", s);if (s[0] == 'C'){inone(x); ++x;inone(a[x]);change(1, 1, n, x);}else{inone(x);int ans = 0;for (auto i : f[1]){if ((i.ft.ft^x) >> (10 - i.ft.sd)) continue;(ans += 1LL * i.sd * inv(1 << (10 - i.ft.sd)) % mod) %= mod;}printf("%d\n", ans);}}}return 0;}