# BZOJ2134: 单选错位

假设当前在做 i 题目，如果a[i+1]>=a[i],那么我们只需要让i+1题目的答案是i的答案即可，ans+=1/a[i+1]

否则 i 题目的答案必须在1--a[i+1]，所以ans+=a[i+1]/a[i]*1/a[i+1]=1/a[i]

换句话说 ans+=min(1/a[i+1],1/a[i])

 1 #include<cstdio>
2
3 #include<cstdlib>
4
5 #include<cmath>
6
7 #include<cstring>
8
9 #include<algorithm>
10
11 #include<iostream>
12
13 #include<vector>
14
15 #include<map>
16
17 #include<set>
18
19 #include<queue>
20
21 #include<string>
22
23 #define inf 1000000000
24
25 #define maxn 10000000+5
26
27 #define maxm 500+100
28
29 #define eps 1e-10
30
31 #define ll long long
32
33 #define pa pair<int,int>
34
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42
43 #define mod 100000001
44
45 using namespace std;
46
48
49 {
50
51     ll x=0,f=1;char ch=getchar();
52
53     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
54
55     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
56
57     return x*f;
58
59 }
60 ll n,a,b,c,d[maxn];
61
62 int main()
63
64 {
65
66     freopen("input.txt","r",stdin);
67
68     freopen("output.txt","w",stdout);
69
71     for2(i,2,n)d[i]=(d[i-1]*a+b)%mod;
72     for1(i,n)d[i]=(d[i]%c)+1;d[n+1]=d[1];
73     double ans=0;
74     for1(i,n)
75     if(d[i+1]>=d[i])ans+=1.0/(double)d[i+1];
76     else ans+=1.0/(double)d[i];
77     printf("%.3f\n",ans);
78
79     return 0;
80
81 }
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posted @ 2014-11-11 16:12  ZYF-ZYF  Views(184)  Comments(0Edit  收藏  举报