Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

You are given a sequence of N integers (each within the range [0, 2¹⁶ - 1] ) along with P operations and in order to solve this problem you need to process the operations instructed as follows.

There are two kinds of operations that you will be instructed to perform:

1) Modification

- Given a non-negative number T , you need to increase the value of every number in the sequence by T . If the value of any number in the sequence is larger than 2¹⁶ - 1 after the operation, you should divide its value by 2¹⁶ and take the remainder as its value;

2) Query

- Given a non-negative number T , query how many numbers in the sequence satisfies the condition that its bitwise and result with 2^T is greater than zero.

For simplicity, all you need to do here is to output the sum ( sum < 10, 000, 000, 000 ) of the answers to all queries.

Input 

There are multiple test cases in the input file. Each test case starts with one integer N (N10⁵) , the number of integers in the sequence. The following N line consists of one integer P (0P2¹⁶ - 1) , the value on i -th line being the value of the i -th number in the sequence.

Each of the following lines is either of the format ``C delta " (0delta) , meaning that you should increase the value of every number by delta, or ``Q i " (0i15) , meaning that you should calculate the answer to the query (as explained in the problem description). Every test case ends with one character `E' on a single line, followed by a blank line.

N = - 1 indicates the end of input file and should not be processed by your program. It is guaranteed that the total number of operations in each test case does not exceed 200,000.

Output 

For each test case, print the sum of answers to queries on one separate line in the format as indicated in the sample output.

Sample Input 

3 
1 
2 
4 
Q 1 
Q 2 
C 1 
Q 1 
Q 2
E

-1

Sample Output 

Case 1: 5

---

題目描述：

操作有 2 種
1. 把所有數字加上同一個數字
2. 將所有數字與 2^i 作 and 運算後，有多少非零數字

題目解法：

用一个计数器S, 累加当前所有的delta, 第i个数现在为(A[i] + S) mod 65536.每次询问第i位上为1的数的个数, 等价于判断当前所有的数中 mod 2^(i+1) >= 2^i的数的个数.对读入的N个数预处理, 统计mod 2i = j的数的个数, 设Sum[i][j]表示mod 2^i <= j的数的个数.而每次询问 2^i <= (x + S) mod 2^(i+1) <= 2^(i+1)-1的x的个数.而这样的x mod 2^i也一定对应一个连续的区间, 利用Sum[i+1]值可以O(1)回答.预处理的时间复杂度为n*16, 每次询问为O(1).

#include <stdio.h>
#include <string.h>
using namespace std;
int table[16][65536];
int main() {
    int n, cases = 0;
    char cmd[10];
    while(scanf("%d", &n) == 1 && n != -1) {
        memset(table, 0, sizeof(table));
        long long ret = 0;
        int i, j, k, P, T;
        for(i = 0; i < n; i++) {
            scanf("%d", &P);
            for(j = 0; j < 16; j++)
                table[j][P%(1<<(j+1))]++;
        }
        for(i = 0; i < 16; i++) {
            for(j = 1; j < 65536; j++)
                table[i][j] += table[i][j-1];
        }
        int S = 0;
        while(scanf("%s", cmd) == 1) {
            if(cmd[0] == 'E')   break;
            scanf("%d", &T);
            if(cmd[0] == 'C') {
                S += T;
                S %= 65536;
            } else {
                i = 1<<T, j = 1<<(T+1);
                int sub = 0;// 2^i <= (x+S) mod 2^(i+1) < 2^(i+1)
                int l = ((i-S)%j+j)%j, r = j-1;
                if(l <= i) {
                    sub = table[T][l+i-1]- (l ? table[T][l-1] : 0);
                } else {
                    sub = table[T][r] - (l ? table[T][l-1] : 0);
                    sub += table[T][(l+i-1)%j];
                }
                ret += sub;
            }
        }
        printf("Case %d: %lld\n", ++cases, ret);
    }
    return 0;
}