HDU 6304 Chiaki Sequence Revisited

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=6304

多校contest1

Problem Description

Chiaki is interested in an infinite sequence a1,a2,a3,..., which is defined as follows:

a n = {1 a n - a n - 1 + a n - 1 - a n - 2 n = 1, 2 n \geq 3

Chiaki would like to know the sum of the first n terms of the sequence, i.e. ∑i=1nai. As this number may be very large, Chiaki is only interested in its remainder modulo (109+7).

Input

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).

Output

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

题目大意是求一个奇怪序列的前 n 项和，n最坏情况达 1e18。

开始打了个表，发现在序列是从1开始的连续整数(每个整数出现的次数不同)，除了数字 1是出现两次，其他数字如x 都是出现 lowbit(x)后缀0的个数+1 次。如2(10) 出现2次，3(11) 出现1次，4(100) 出现3次，5(101)出现1次，6(110) 出现2次...

可以发现这是一个十分有特点的(类似2进制，有部分对称性)的序列

接下来我们可以发现如果把1当做出现1次，出现在 2^n 上，如果占满，次数总和刚好是 2^(n+1) -1 次，那么多出来的数似乎又没有规律了，这时我们可以利用局部对称与这个和二进制相似的特点，找到第 n 个数是数字多少(落在图中的位置)，

有两种方法:

第一种可以知道N(自减一后)对应的准确数字，但不知N落在的数字差几次被填满，不便计算。

 1 void init(){
 2     P[0]=1;P[1]=2;
 3     nP[0]=0;nP[1]=1;
 4     for(int i=2;i<=62;++i){
 5         P[i]=2*P[i-1];
 6         nP[i]=P[i]-1;
 7     }
 8 }
 9 
10 ll getbound(ll N){
11     ll bound=0;
12     for(int i=62;i>=1;--i){
13         while(N>=nP[i]){
14             N-=nP[i];
15             bound+=P[i-1];
16         }
17     }
18     return bound;
19 }

第二种完全参照二进制，可知N(自减一后)所落在的数字最近的次数填满数字，后来计算时很方便。

 1 void init(){
 2     P[0]=1;P[1]=2;
 3     nP[0]=0;nP[1]=1;
 4     for(int i=2;i<=62;++i){
 5         P[i]=2*P[i-1];
 6         nP[i]=P[i]-1;
 7     }
 8 }
 9 
10 ll getbound(ll& N){
11     ll bound=0;
12     for(int i=62;i>=1;--i){
13         if(N>=nP[i]){
14             N-=nP[i];
15             bound+=P[i-1];
16         }
17     }
18     return bound;
19 }

计算时可以可利用每个数字出现的次数，是1(2^0)的倍数的出现过一次，是2(2^1)的倍数的额外出现过一次，是4(2^2)的倍数的又额外出现一次，，，(这也恰恰是后缀0的意义)

在这里贴两份按照上述两种方法写的代码。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long ll;
 5 
 6 const ll MOD=1e9+7;
 7 ll P[66];
 8 ll nP[66];
 9 ll a[101];
10 ll arr[101];
11 
12 ll lowbit(ll x){
13     ll low=x&(-x);
14     ll cnt=0;
15     while(low>>=1){
16         cnt++;
17     }
18     return cnt;
19 }
20 
21 void init(){
22     P[0]=1;P[1]=2;
23     nP[0]=0;nP[1]=1;
24     for(int i=2;i<=62;++i){
25         P[i]=2*P[i-1];
26         nP[i]=P[i]-1;
27     }
28     
29     //a[1]=1;a[2]=1;
30     //arr[1]=1;arr[2]=2;
31     //for(int i=3;i<101;i++){
32     //    a[i]=a[i-a[i-1]]+a[i-1-a[i-2]];
33     //    arr[i]=arr[i-1]+a[i];
34     //}
35     
36     //for(int i=1;i<101;++i) printf("%lld\n",arr[i]);
37     
38 }
39 
40 ll inv(ll a,ll m){
41     if(a==1) return 1;
42     return inv(m%a,m)*(m-m/a)%m;
43 }
44 
45 ll getbound(ll N){
46     ll bound=0;
47     for(int i=62;i>=1;--i){
48         while(N>=nP[i]){
49             N-=nP[i];
50             bound+=P[i-1];
51         }
52     }
53     return bound;
54 }
55 
56 int main(){
57     //freopen("data.in","r",stdin);
58     //freopen("data1.out","w",stdout);
59     init();
60     int T;
61     scanf("%d",&T);
62     while(T--){
63         ll N;
64         scanf("%lld",&N);
65         if(N==1) puts("1");
66         else{
67             ll ans=0;
68             N-=1ll;
69             ll bound=0;
70             bound=getbound(N);
71             //printf("bound = %lld\n",bound);
72             ll cnt=lowbit(bound)+1ll;
73             ll tot=N;
74             for(ll i=1;i<=cnt;++i){
75                 if(bound==getbound(N+i)) tot++;
76                 else break;
77             }
78             ll _m2=inv(2,MOD);
79             //printf("%lld\n",_m2);
80             for(int i=0;i<=62;++i){
81                 if(P[i]<=bound){
82                     ll M=bound/P[i];
83                     ans=(ans+(P[i]%MOD)*(M%MOD)%MOD*((M+1ll)%MOD)%MOD*_m2%MOD)%MOD;
84                 }
85                 else break;
86             }
87             //printf("1:%lld\n",(ans+1)%MOD);
88             ans=(ans-(bound)*(tot-N)%MOD+MOD)%MOD;
89             printf("%lld\n",(ans+1ll)%MOD);
90             //printf("%I64d %I64d\n",(ans+1ll)%MOD,arr[N+1]);
91             //最后加一 
92         }
93     }
94     return 0;
95 }

View Code

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long ll;
 5 
 6 const ll MOD=1e9+7;
 7 ll P[66];
 8 ll nP[66];
 9 ll a[101];
10 ll arr[101];
11 
12 ll lowbit(ll x){
13     ll low=x&(-x);
14     ll cnt=0;
15     while(low>>=1){
16         cnt++;
17     }
18     return cnt;
19 }
20 
21 void init(){
22     P[0]=1;P[1]=2;
23     nP[0]=0;nP[1]=1;
24     for(int i=2;i<=62;++i){
25         P[i]=2*P[i-1];
26         nP[i]=P[i]-1;
27     }
28     
29     //a[1]=1;a[2]=1;
30     //arr[1]=1;arr[2]=2;
31     //for(int i=3;i<101;i++){
32     //    a[i]=a[i-a[i-1]]+a[i-1-a[i-2]];
33     //    arr[i]=arr[i-1]+a[i];
34     //}
35     
36     //for(int i=1;i<101;++i) printf("%lld\n",arr[i]);
37     
38 }
39 
40 ll inv(ll a,ll m){
41     if(a==1) return 1;
42     return inv(m%a,m)*(m-m/a)%m;
43 }
44 
45 ll getbound(ll& N){
46     ll bound=0;
47     for(int i=62;i>=1;--i){
48         if(N>=nP[i]){
49             N-=nP[i];
50             bound+=P[i-1];
51         }
52     }
53     return bound;
54 }
55 
56 int main(){
57     //freopen("data.in","r",stdin);
58     //freopen("data1.out","w",stdout);
59     init();
60     int T;
61     scanf("%d",&T);
62     while(T--){
63         ll N;
64         scanf("%lld",&N);
65         if(N==1) puts("1");
66         else{
67             ll ans=0;
68             N-=1ll;
69             ll bound=0;
70             bound=getbound(N);
71             //printf("bound = %lld\n",bound);
72             //ll cnt=lowbit(bound)+1ll;
73             //ll tot=N;
74             //for(ll i=1;i<=cnt;++i){
75             //    if(bound==getbound(N+i)) tot++;
76             //    else break;
77             //}
78             ll _m2=inv(2,MOD);
79             for(int i=0;i<=62;++i){
80                 if(P[i]<=bound){
81                     ll M=bound/P[i];
82                     ans=(ans+(P[i]%MOD)*(M%MOD)%MOD*((M+1ll)%MOD)%MOD*_m2%MOD)%MOD;
83                     
84                 }
85                 else break;
86             }
87             //printf("1:%lld\n",(ans+1)%MOD);
88             ans=(ans+(bound+1)*N%MOD)%MOD;
89             printf("%lld\n",(ans+1ll)%MOD);
90             //printf("%I64d %I64d\n",(ans+1ll)%MOD,arr[N+1]);
91             //最后加一 
92         }
93     }
94     return 0;
95 }

View Code

然后，比赛时WA了两发，其实规律找到了，但错在了计算，算总和时，第一个错误处是没用逆元，第二个错误处是P[i]在N为1e18时奇大，应该先mod在相乘。

血的教训。

Chiaki is interested in an infinite sequence a1,a2,a3,..., which is defined as follows:

a n = {1 a n - a n - 1 + a n - 1 - a n - 2 n = 1, 2 n \geq 3

Chiaki would like to know the sum of the first n terms of the sequence, i.e. ∑i=1nai. As this number may be very large, Chiaki is only interested in its remainder modulo (109+7).

posted on 2018-07-24 22:11 Emiya_Kiritsugu 阅读(332) 评论(0) 编辑收藏举报

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HDU 6304 Chiaki Sequence Revisited

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