FZU 2105 Digits Count（位数计算）

Description

题目描述

Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are integers. For L≤i≤R, we do A[i]=A[i] AND opn (here "AND" is bitwise operation). Operation 2: OR opn L R Here opn, L and R are integers. For L≤i≤R, we do A[i]=A[i] OR opn (here "OR" is bitwise operation). Operation 3: XOR opn L R Here opn, L and R are integers. For L≤i≤R, we do A[i]=A[i] XOR opn (here "XOR" is bitwise operation). Operation 4: SUM L R We want to know the result of A[L]+A[L+1]+...+A[R]. Now can you solve this easy problem?

给定N个整数A={A[0],A[1],...,A[N-1]}。下面有一些操作。 操作1：AND opn L R 其中opn，L和R都是整数。 对于L≤i≤R，我们使A[i]=A[i] AND opn（这里的"AND"是位运算）。 操作2：OR opn L R 其中opn，L和R都是整数。 对于L≤i≤R，我们使A[i]=A[i] OR opn（这里的"OR"是位运算）。 操作3：XOR opn L R 其中opn，L和R都是整数。 对于L≤i≤R，我们使A[i]=A[i] XOR opn（这里的"XOR"是位运算）。 操作4：SUM L R 我们想要知道A[L]+A[L+1]+...+A[R]的值。 现在，你能解决这个简单的问题吗？

Input

输入

The first line of the input contains an integer T, indicating the number of test cases. (T≤100) Then T cases, for any case, the first line has two integers n and m (1≤n≤1,000,000, 1≤m≤100,000), indicating the number of elements in A and the number of operations. Then one line follows n integers A[0], A[1], ..., A[n-1] (0≤A[i]<16,0≤i<n). Then m lines, each line must be one of the 4 operations above. (0≤opn≤15)

第一行有一个整数T，表示测试样例的数量。(T≤100) 后面有T个样例，每个样例的第一行有2个整数n和m(1≤n≤1,000,000, 1≤m≤100,000)，表示元素数量与操作次数。 下一行有n个整数A[0], A[1], ..., A[n-1] (0≤A[i]<16,0≤i<n)。 接着m行，每行都是上述4种操作的其中一个。(0≤opn≤15)

Output	输出
For each test case and for each "SUM" operation, please output the result with a single line.	对于每个测试样例的每个"SUM"操作，都要输出到单独的一行。

Sample Input - 输入样例	Sample Output - 输出样例
1 4 4 1 2 4 7 SUM 0 2 XOR 5 0 0 OR 6 0 3 SUM 0 2	7 18

【题解】

　　单纯的线段树。

　　利用线段树查找快速的特点保存元素，对于某段区间内相同状态的元素进行合并。

　　进行操作的时候不断查找，直到可操作的区间。因为数据都是非负整数，所以可以用负数标记元素不同的区间。

　　PS：状态发生变化的时候都要下压，注意深入的方向。

　　PS2：简单的运算速度是高于开辟空间的速度，所以没必要把线段树每个节点的左右区间都记录下来。

　　PSP：开辟元素数量4倍的空间一定够用。

【代码 C++】

 #include<cstdio>
 int data[ << ], opn, L, R;
 char ts[];
 int build(int data_i, int L, int R){//[L, R]
     if (L == R) scanf("%d", &data[data_i]);
     else{
         int mid = (L + R) >> ;// [L, mid]  (mid, R]
         L = build(data_i <<  | , L, mid);
         R = build(data_i +  << , mid + , R);
         if (L == R) data[data_i] = L;
         else data[data_i] = -;
     }
     return data[data_i];
 }
 int bitwise_peration(int data_i, int L_now, int R_now){
     if (data[data_i] != - && L <= L_now && R_now <= R){
         if (*ts == 'A') data[data_i] &= opn;
         else if (*ts == 'O') data[data_i] |= opn;
         else if (*ts == 'X') data[data_i] ^= opn;
     }
     else{
         if (data[data_i] != -) data[data_i <<  | ] = data[data_i +  << ] = data[data_i];
         int mid = (R_now + L_now) >> ;
         if (R <= mid){// goto Right
             L_now = bitwise_peration(data_i <<  | , L_now, mid);
             R_now = data[data_i +  << ];
         }
         else if (mid < L){// goto Left
             L_now = data[data_i <<  | ];
             R_now = bitwise_peration(data_i +  << , mid + , R_now);
         }
         else{
             L_now = bitwise_peration(data_i <<  | , L_now, mid);
             R_now = bitwise_peration(data_i +  << , mid + , R_now);
         }
         if (L_now == R_now) data[data_i] = L_now;
         else data[data_i] = -;
     }
     return data[data_i];
 }
 int sum(int data_i, int L_now, int R_now){
     if (data[data_i] != - && L <= L_now && R_now <= R){
         return data[data_i] * (R_now - L_now + );
     }
     if (data[data_i] != -) data[data_i <<  | ] = data[data_i +  << ] = data[data_i];
     int mid = (R_now + L_now) >> ;
     if (R <= mid) return sum(data_i <<  | , L_now, mid);
     if (mid < L) return sum(data_i +  << , mid + , R_now);
     return sum(data_i <<  | , L_now, mid) + sum(data_i +  << , mid + , R_now);
 }
 int main(){
     int t, m, n, i;
     for (scanf("%d", &t); t; --t){
         scanf("%d%d", &n, &m);
         --n;
         build(, , n);
         while (m--){
             scanf("%s", ts);
             if (*ts == 'S'){
                 scanf("%d%d", &L, &R);
                 printf("%d\n", sum(, , n));
             }
             else{
                 scanf("%d%d%d", &opn, &L, &R);
                 bitwise_peration(, , n);
             }
         }
     }
     return ;
 }

 #include<cstdio>
 int L, R;
 char data[], opn, ts[];
 int read_g(){// 输入挂
     int add = getchar() - '';
     int a = getchar();
     while (a >= '' && a <= '') add = add *  + a - '', a = getchar();
     return add;
 }
 int build(int data_i, int L, int R){// [L, R]
     if (L == R) data[data_i] = read_g();
     else{// [L, mid]  (mid, R]
         int mid = (L + R) >> ;
         L = build(data_i <<  | , L, mid);
         R = build(data_i +  << , ++mid, R);
         if (L == R) data[data_i] = L;
         else data[data_i] = -;
     }
     return data[data_i];
 }
 int bitwise_peration(int data_i, int L_now, int R_now){
     if (~data[data_i] && L <= L_now && R_now <= R){
         if (*ts == 'A') data[data_i] &= opn;
         else if (*ts == 'O') data[data_i] |= opn;
         else if (*ts == 'X') data[data_i] ^= opn;
     }
     else{
         if (~data[data_i]) data[data_i <<  | ] = data[data_i +  << ] = data[data_i];
         int mid = (R_now + L_now) >> ;
         if (R <= mid){// goto Right
             L_now = bitwise_peration(data_i <<  | , L_now, mid);
             R_now = data[data_i +  << ];
         }
         else if (mid < L){// goto Left
             L_now = data[data_i <<  | ];
             R_now = bitwise_peration(data_i +  << , mid + , R_now);
         }
         else{
             L_now = bitwise_peration(data_i <<  | , L_now, mid);
             R_now = bitwise_peration(data_i +  << , mid + , R_now);
         }
         if (L_now == R_now) data[data_i] = L_now;
         else data[data_i] = -;
     }
     return data[data_i];
 }
 int sum(int data_i, int L_now, int R_now){
     if (~data[data_i]){
         if (L_now < L) L_now = L;
         if (R < R_now) R_now = R;
         return data[data_i] * (++R_now - L_now);
     }
     int mid = (R_now + L_now) >> ;
     if (R <= mid) return sum(data_i <<  | , L_now, mid);
     if (mid < L) return sum(++data_i << , ++mid, R_now);
     return sum(data_i <<  | , L_now, mid) + sum(data_i +  << , mid + , R_now);
 }
 int main(){
     int m, n;
     short t = read_g();
     while (t--){
         n = read_g(); m = read_g(); --n;
         build(, , n);
         while (m--){
             scanf("%s", ts); getchar();
             if (*ts == 'S'){
                 L = read_g(); R = read_g();
                 printf("%d\n", sum(, , n));
             }
             else{
                 opn = read_g(); L = read_g(); R = read_g();
                 bitwise_peration(, , n);
             }
         }
     }
     return ;
 }

FZU 2105