BZOJ2134: 单选错位
题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2134
题解：因为每个答案之间是互不影响的，所以我们可以挨个计算。
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])
代码：
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 47 inline ll read() 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 70 n=read();a=read();b=read();c=read();d[1]=read(); 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 }
