牛客网暑期ACM多校训练营(第三场): C. Chiaki Sequence Reloaded（数位DP）

题目描述

Chiaki is interested in an infinite sequence a1, a2, a3, ..., which defined as follows:
$a_n=\begin{cases}0 & n = 1 \\ a_{\lfloor \frac{n}{2} \rfloor} + (-1)^{\frac{n(n+1)}{2}} & n \ge 2\end{cases}$
Chiaki would like to know the sum of the first n terms of the sequence, i.e. $\sum\limits_{i=1}^{n}|a_i|$ . As this number may be very large, Chiaki is only interested in its remainder modulo (109 + 7).

输入描述:

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 105), indicating the number of test cases. For each test case:
The first line contains an integer n (1 ≤ n ≤ 1018).

输出描述:

For each test case, output an integer denoting the answer.

题意：求上面的公式的绝对值的前缀和

打表可以看出a[i] = （i的2进制表示下：相邻位相同的个数 - 相邻位不同的个数），例如a[13(1101)] = 1-2 = -1

然后就可以求数位DP了，因为i的2进制最高只有64位，所以a[i]的取值范围大概是[-64, 64]，可以考虑暴力每个值各出现多少次

预处理sum[x][y][z]表示当前暴力到第x位，且第x位为y，有多少个数满足a[]值为z（包含前导0）

例如sum[5][1][-3]就表示刚好有5位，且最高位为1，且满足a[k]为-3的二进制k的个数，有转移

for(i=2;i<=66;i++)
{for(j=1;j<=135;j++)     //因为数组下标不能为负数，所以需要整体移位：[-64,64]→[1,129]{sum[i][0][j] = (sum[i-1][0][j-1]+sum[i-1][1][j+1])%mod;sum[i][1][j] = (sum[i-1][1][j-1]+sum[i-1][0][j+1])%mod;}
}

之后对于每一个询问n，从n的最高位开始，如果当前位为1，那么很显然将这一位改成0之后后面所有的数字都可以01任取，与此同时你可以算出前面的（二进制相邻位相同的个数 - 相邻位不同的个数），这样只需要算上你预处理的sum[]就可以求出当前位对答案的贡献了，不过直接暴力sum[]还是会超时，所以还需要对sum再求一次前缀和

超时但好理解的程序：

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define LL long long
LL sum[88][2][222], a[88];
int main(void)
{LL T, n, i, j, len, ans, now, p;sum[1][0][100] = sum[1][1][100] = 1;for(i=2;i<=66;i++){for(j=1;j<=200;j++){sum[i][0][j] = sum[i-1][0][j-1]+sum[i-1][1][j+1];sum[i][1][j] = sum[i-1][1][j-1]+sum[i-1][0][j+1];}}scanf("%lld", &T);while(T--){len = 0;scanf("%lld", &n);while(n){a[++len] = n%2;n /= 2;}ans = now = 0;for(i=len;i>=1;i--){p = 0;if(i!=len && a[i]!=a[i+1])  p = -1;else  if(i!=len) p = 1;if(a[i]){if(i!=len){for(j=1;j<=200;j++)ans += sum[i][0][j]*abs(j-100+now-p);}}if(i!=len){for(j=1;j<=200;j++)ans += sum[i][1][j]*abs(j-100);}now += p;}ans += abs(now);printf("%lld\n", ans);}return 0;
}

AC代码（附打表程序）：

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define LL long long
#define mod 1000000007
LL sum[70][2][138], Q[70][2][138], a[88];
int main(void)
{LL T, n, i, j, k, len, ans, now, p;sum[1][0][65] = sum[1][1][65] = 1;for(i=2;i<=66;i++){for(j=1;j<=135;j++)        //因为数组下标不能为负数，所以需要整体移位：[-64,64]→[1,129]{sum[i][0][j] = (sum[i-1][0][j-1]+sum[i-1][1][j+1])%mod;sum[i][1][j] = (sum[i-1][1][j-1]+sum[i-1][0][j+1])%mod;}}for(i=1;i<=66;i++){for(k=1;k<=135;k++){for(j=1;j<=135;j++){Q[i][0][k] = (Q[i][0][k]+sum[i][0][j]*abs(j-k))%mod;Q[i][1][k] = (Q[i][1][k]+sum[i][1][j]*abs(j-k))%mod;}}}scanf("%lld", &T);while(T--){len = 0;scanf("%lld", &n);while(n){a[++len] = n%2;n /= 2;}ans = now = 0;for(i=len;i>=1;i--){p = 0;if(i!=len && a[i]!=a[i+1])  p = -1;else  if(i!=len) p = 1;if(a[i]){if(i!=len)ans = (ans+Q[i][0][65-now+p])%mod;//sum[i][0][j]*abs(j-65+now-p);}if(i!=len)ans = (ans+Q[i][1][65])%mod;now += p;}ans = (ans+abs(now))%mod;printf("%lld\n", ans);}return 0;
}
/*#include<stdio.h>
#include<stdlib.h>
int F[10005], S[10005];
void Jud(int x)
{if(x<=1)printf("%d", x);else{Jud(x/2);printf("%d", x%2);}
}
int main(void)
{int i;for(i=1;i<=9999;i++){if(i==1)F[i] = 0;else{if(i%4==2 || i%4==1)F[i] = F[i/2]-1;elseF[i] = F[i/2]+1;}//printf("%-5d", i);//Jud(i);//printf(":   %d\n", F[i]);S[i] = abs(F[i])+S[i-1];if(i<=15 || i==7788)printf("%d\n", S[i]);}return 0;
}*/