TopCoder 11379 YetAnotherORProblem / 模拟赛 T3 yet another 计数题

原题：

Problem Statement

NOTE: This problem statement contains subscripts that may not display properly if viewed outside of the applet. You’re given a long[] R containing N elements. Count the number of sequences A0, A1, …, AN-1 such that each Ai is an integer satisfying 0 &le Ai &le R[i] and A0 + A1 + … + AN-1 = A0 | A1 | … | AN-1. The ‘|’ symbol stands for bitwise OR of the operands. Return the number of such sequences modulo 1,000,000,009.

Definition

Class: YetAnotherORProblem
Method: countSequences
Parameters: long[]
Returns: int
Method signature: int countSequences(long[] R)
(be sure your method is public)

Notes

• If a and b are single bits then a | b is defined as max(a, b). For two integers, A and B, in order to calculate A | B, they need to be represented in binary: A = (an…a1)2, B = (bn…b1)2 (if the lengths of their representations are different, the shorter one is prepended with the necessary number of leading zeroes). Then A | B = C = (cn…c1)2, where ci = ai | bi. For example, 10 | 3 = (1010)2 | (0011)2 = (1011)2 = 11.

Constraints

• R will contain between 2 and 10 elements, inclusive.
• Each element of R will be between 1 and 10^18, inclusive.

Examples

0）
{3,5}
Returns: 15
All the possible sequences are: {0,0}, {0,1}, {0,2}, {0,3}, {0,4}, {0,5}, {1,0}, {1,2}, {1,4}, {2,0}, {2,1}, {2,4}, {2,5}, {3,0}, {3,4}.
1)
{3,3,3}
Returns: 16
2)
{1,128}
Returns: 194
3)
{26,74,25,30}
Returns: 8409
4)
{1000000000,1000000000}
Returns: 420352509

模拟赛题（翻译）：

给你一个数组R，包含N个元素，求有多少满足条件的序列AA使得

0 ≤ $A[i]$ ≤ $R[i]$
$A[0]$+$A[1]$+…+$A[N-1]$ = $A[0]$ | $A[1]$… | $A[N-1]$
输出答案对1e9+9取模

输入

第一行输入一个整数N (102 ≤ N ≤ 10)

第二行输入N个整数 $R[i]$ (1 ≤ $R[i]$ ≤ 1e18)

输出

输出一个整数

样例

输入
2
3 5
输出
15
输入
3
3 3 3
输出
16
输入
2
1000000000 1000000000
输出
420352509
提示
子任务1: n <= 3 && max(R[i]) <= 100
子任务2: n <= 5 && max(R[i]) <= 1000
子任务3: 无限制

注: 题目在输入是有所不同的。

题解：

我们知道，在二进制数中，
和如果和等于或的话
那么只有两种情况：

(1) 多个数该二进制位的只有一个1.

(2) 多个数该二进制位的都没有1.

我们可以考虑数位dp
假设$dp[i][mask]$ 表示已考虑i位之前的情况，当前的限制情况为mask.(其中第i为的1 或 0表示第 i 个数(从0开始算) 是否有限制，我们假定1为有限制，0 为没限制).
这题细节很多。
我们在代码中继续讲。topcoder上只需按照题目要求打包成一个类即可。

代码如下：

#include 
#include 
#include 
#include 
using namespace std;
long long r[20];
const int mod = 1e9 + 9;
long long dp[65][1 << 11];
int n;
long long dfs(int w, int lim){//记忆化搜索,w表示当前第几位,lim是被限制的状态
	if(w < 0) return 1;//有一组A满足要求
	if(dp[w][lim] != -1) return dp[w][lim];
	int i, j, k;
	int now = 0;
	long long res = 0;
	//先考虑为一的情况
	for(i = 0; i < n; i++){ //枚举哪一个数该位为一,其他位为0
		now = 0;
		if((lim & (1 << i)) &&  !(r[i] & (1LL << w))) continue; //不可能为1(可能比较绕),就是有限制并且该二进制位最多为0
		for(j = 0; j < n; j++){//枚举每个数之后的限制情况
			if(i == j){//枚举到该位为一的数
				if(lim & (1 << i)){//注意特判,假如当前无限制，后面不可能有限制
					now |= (1 << j);
				}
			}
			else if((!(r[j] & (1LL << w))) && (lim & (1 << j))){//枚举到其他数,当前有限制，并且最高只能取0,后面任然会有限制
				now |= (1 << j);
			}
		}
		res = (res + dfs(w - 1, now)) % mod;
	}
	now = 0;
	for(i = 0; i < n; i++){ //枚举一位全为0
		if(!(r[i] & (1LL << w)) && (lim & (1 << i)))//当前有限制，并且最高只能取0,后面任然会有限制
			now |= (1 << i);
	}
	res = (res + dfs(w - 1, now)) % mod;//继续下一位
	dp[w][lim] = res;
	return res;
}
int main(){
	int i;
	scanf("%d", &n);
	for(i = 0; i < n; i++){
		scanf("%lld", &r[i]);
	}
	memset(dp, -1, sizeof(dp));
	printf("%lld\n", dfs(62, (1 << n) - 1));
	return 0;
}