Miaomiao's Function
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 79 Accepted Submission(s): 18
Problem Description
Firstly,Miaomiao define two functions f(x),g(x):
(K is the smallest integer which satisfied x + 9 * K > 0)
if e.g. g(178) = 1 - 7 + 8 = 2,g(1) = 1,g(1234) = 1 - 2 + 3 - 4 = -2;
For example f(20140810) = f(2 + 0 + 1 + 4 + 0 + 8 + 1 + 0) = f(16) = f(1 + 6) = f(7) = 7
Then,you are given two integers L,R( L <= R) .
Answer is defined as .
Please caculate (Answer Mod f(Answer) + f(Answer)) Mod f(Answer).
Pay attantion ! If f(Answer) = 0,please output "Error!"
(K is the smallest integer which satisfied x + 9 * K > 0)
if e.g. g(178) = 1 - 7 + 8 = 2,g(1) = 1,g(1234) = 1 - 2 + 3 - 4 = -2;
For example f(20140810) = f(2 + 0 + 1 + 4 + 0 + 8 + 1 + 0) = f(16) = f(1 + 6) = f(7) = 7
Then,you are given two integers L,R( L <= R) .
Answer is defined as .
Please caculate (Answer Mod f(Answer) + f(Answer)) Mod f(Answer).
Pay attantion ! If f(Answer) = 0,please output "Error!"
Input
There are many test cases. In the first line there is a number T ( T <= 50 ) shows the number of test cases.
For each test cases there are two numbers L,R ( 0 <= L,R <= 10^100 ).
For your information,L,R don't have leading zeros.
For each test cases there are two numbers L,R ( 0 <= L,R <= 10^100 ).
For your information,L,R don't have leading zeros.
Output
For each opeartion,output the result.
Sample Input
2 0 0 0 21
Sample Output
Error! 1HintThe Answer is 0 and 13. So the result is Error! and 13 Mod (1 + 3) = 1
题源为 bestcoder #4 C题
题目大意
给出L,R,询问[L,R]之中所有数的交错和之和,交错和的定义为
给定一个数 x,设它十进制展从高位到低位上的数位依次是 a0, a1, ..., an - 1,定义交错和函数:
f(x) = a0 - a1 + a2 - ... + ( - 1)n - 1an - 1
其实就是hihocoder 1033的题目并将ans做一个模处理
解题思路
这一题的方法和hihocoder 1033完全一致,推不出公式就写了记忆化dfs+java大数……
注意最后的结果可能有0和9两种 所以直接对九取模就会造成答案模数的Error变为0
特判即可
code:
import java.util.*; import java.math.*; public class Main { static class node { public BigInteger s=BigInteger.ZERO,n=BigInteger.ZERO; } public static int top=0,bn=0; public static node[] dp=new node[105]; public static int[] bt=new int[105]; public static node dfs(int pos,int limit) { node tmp = new node(); tmp.s=BigInteger.ZERO; tmp.n=BigInteger.ZERO; if (pos==0) { tmp.s=BigInteger.ZERO; tmp.n=BigInteger.ONE; return tmp; } if (limit==0&&(!dp[pos].n.equals(BigInteger.ZERO))) return dp[pos]; int head=(pos==top)?1:0; int tail=(limit==1)?bt[pos]:9; for (int i=head;i<=tail;i++) { node s= dfs(pos-1,(limit==1&&bt[pos]==i)?1:0); tmp.n=tmp.n.add(s.n); tmp.s=tmp.s.add(s.s); if ((top-pos)%2==0) tmp.s=tmp.s.add((s.n).multiply(BigInteger.valueOf(i))); else tmp.s=tmp.s.add((s.n).multiply(BigInteger.valueOf(-i))); } if (limit==0) dp[pos]=tmp; return tmp; } public static BigInteger f(BigInteger x) { BigInteger ans=BigInteger.ZERO; if (x.equals(BigInteger.ZERO)) return ans; if (x.equals(BigInteger.valueOf(-1))) return ans; bn=0; while (!x.equals(BigInteger.ZERO)) { bn++; bt[bn]=x.mod(BigInteger.valueOf(10)).intValue(); x=x.divide(BigInteger.valueOf(10)); } for (top=1;top<=bn;top++) { for (int i=1;i<=104;i++){ dp[i]=new node(); dp[i].s=BigInteger.ZERO; dp[i].n=BigInteger.ZERO; } ans=ans.add(dfs(top,(top==bn)?1:0).s); } return ans; } public static BigInteger g(BigInteger x) { BigInteger t=BigInteger.ZERO; BigInteger p=x; if (x.equals(BigInteger.ZERO)) return x; if (x.compareto(BigInteger.valueOf(0))<0) if (x.mod(BigInteger.valueOf(9)).equals((BigInteger.valueOf(0)))) return(BigInteger.valueOf(9)); else return (g(x.multiply(BigInteger.valueOf(-8)))); if (x.mod(BigInteger.valueOf(9)).equals(BigInteger.ZERO)) return BigInteger.valueOf(9); else return x.mod(BigInteger.valueOf(9)); } public static void main(String [] args) { Scanner in=new Scanner(system.in); int T=in.nextInt(); while (T--!=0) { BigInteger L=in.nextBigInteger(); BigInteger R=in.nextBigInteger(); BigInteger ans=f(R).add(f(L.add((BigInteger.valueOf(-1)))).negate()); BigInteger t=g(ans); if (t.equals(BigInteger.valueOf(0))) System.out.println("Error!"); else System.out.println((ans.mod(t).add(t)).mod(t)); } } }
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