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HDU 5920 Ugly Problem 【模拟】 (2016中国大学生程序设计竞赛(长春))

2024-08-15 15:35:20   247人阅读

 

Ugly Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
Special Judge

Problem Description
Everyone hates ugly problems.

You are given a positive integer. You must represent that number by sum of palindromic numbers.

A palindromic number is a positive integer such that if you write out that integer as a string in decimal without leading zeros, the string is an palindrome. For example, 1 is a palindromic number and 10 is not.
 

 

Input
In the first line of input, there is an integer T denoting the number of test cases.

For each test case, there is only one line describing the given integer s (1≤s≤101000).
 

 

Output
For each test case, output “Case #x:” on the first line where x is the number of that test case starting from 1. Then output the number of palindromic numbers you used, n, on one line. n must be no more than 50. en output n lines, each containing one of your palindromic numbers. Their sum must be exactly s.
 

 

Sample Input
2
18
1000000000000
 

 

Sample Output
Case #1:
2
9
9
Case #2:
2
999999999999
1
Hint
9 + 9 = 18
999999999999 + 1 = 1000000000000
 

 

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5920

题目大意：

　　输入一个长整数s（s<=101000），求将其拆分为不超过50个回文串之和的方案。

题目思路：

　　【模拟】

　　将前半段取出来，-1，构造成回文串c，s-=c，直到c=1。

　　特殊处理0~20的情况。

技术分享
  1 //  2 //by coolxxx  3 //#include<bits/stdc++.h>  4 #include<iostream>  5 #include<algorithm>  6 #include<string>  7 #include<iomanip>  8 #include<map>  9 #include<stack> 10 #include<queue> 11 #include<set> 12 #include<bitset> 13 #include<memory.h> 14 #include<time.h> 15 #include<stdio.h> 16 #include<stdlib.h> 17 #include<string.h> 18 //#include<stdbool.h> 19 #include<math.h> 20 #define min(a,b) ((a)<(b)?(a):(b)) 21 #define max(a,b) ((a)>(b)?(a):(b)) 22 #define abs(a) ((a)>0?(a):(-(a))) 23 #define lowbit(a) (a&(-a)) 24 #define sqr(a) ((a)*(a)) 25 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b)) 26 #define mem(a,b) memset(a,b,sizeof(a)) 27 #define eps (1e-10) 28 #define J 10 29 #define mod 1000000007 30 #define MAX 0x7f7f7f7f 31 #define PI 3.14159265358979323 32 #pragma comment(linker,"/STACK:1024000000,1024000000") 33 #define N 2004 34 using namespace std; 35 typedef long long LL; 36 double anss; 37 LL aans,sum; 38 int cas,cass; 39 int n,m,lll,ans; 40 char s[N]; 41 int a[54][N],b[N],c[N]; 42 void gjdjian(int a[],int b[]) 43 { 44     int i; 45     for(i=1;i<=b[0];i++) 46         a[i]-=b[i]; 47     for(i=1;i<=a[0];i++) 48         if(a[i]<0) 49             a[i]+=J,a[i+1]--; 50     while(a[0]>1 && !a[a[0]])a[0]--; 51 } 52 void gjdprint(int a[]) 53 { 54     int i; 55     for(i=a[0];i;i--) 56         printf("%d",a[i]); 57     puts(""); 58 } 59 void print() 60 { 61     int i,j; 62     printf("Case #%d:\n",cass); 63     printf("%d\n",lll); 64     for(i=1;i<=lll;i++) 65         gjdprint(a[i]); 66 } 67 int main() 68 { 69     #ifndef ONLINE_JUDGEW 70     freopen("1.txt","r",stdin); 71 //    freopen("2.txt","w",stdout); 72     #endif 73     int i,j,k; 74 //    init(); 75 //    for(scanf("%d",&cass);cass;cass--) 76     for(scanf("%d",&cas),cass=1;cass<=cas;cass++) 77 //    while(~scanf("%s",s)) 78 //    while(~scanf("%d",&n)) 79     { 80         lll=0;mem(a,0); 81         scanf("%s",s); 82         n=strlen(s); 83         b[0]=n; 84         for(i=0;i<n;i++)b[n-i]=s[i]-‘0‘; 85         while(!(b[0]==1 && b[1]==0)) 86         { 87             if(b[0]==1) 88             { 89                 a[++lll][0]=1; 90                 a[lll][1]=b[1]; 91                 break; 92             } 93             else if(b[0]==2 && b[2]==1) 94             { 95                 if(b[1]==0) 96                 { 97                     a[++lll][0]=1; 98                     a[lll][1]=9; 99                     a[++lll][0]=1;100                     a[lll][1]=1;101                     break;102                 }103                 else if(b[1]==1)104                 {105                     a[++lll][0]=2;106                     a[lll][1]=1;107                     a[lll][2]=1;108                     break;109                 }110                 else111                 {112                     a[++lll][0]=2;113                     a[lll][1]=1;114                     a[lll][2]=1;115                     a[++lll][0]=1;116                     a[lll][1]=b[1]-1;117                     break;118                 }119             }120             else121             {122                 for(i=b[0];i>b[0]/2;i--)123                     c[i-b[0]/2]=b[i];124                 c[0]=(b[0]+1)/2;125                 int d[2]={1,1};126                 gjdjian(c,d);127                 j=c[0]+c[0];128                 while(j>b[0])j--;129                 lll++;130                 a[lll][0]=j;131                 for(i=c[0];i;i--,j--)132                     a[lll][c[0]-i+1]=a[lll][j]=c[i];133                 gjdjian(b,a[lll]);134             }135         }136         print();137     }138     return 0;139 }140 /*141 //142 143 //144 */
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HDU 5920 Ugly Problem 【模拟】 (2016中国大学生程序设计竞赛(长春))


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