标签：
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 JudgeProblem DescriptionEveryone 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.
InputIn 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).
OutputFor 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 Input2181000000000000
Sample OutputCase #1:299Case #2:29999999999991Hint9 + 9 = 18999999999999 + 1 = 1000000000000
Statistic | Submit | Clarifications | Back
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。
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 else 111 { 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 else 121 { 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 */
HDU 5920 Ugly Problem 【模拟】 (2016中国大学生程序设计竞赛(长春))
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原文地址：http://www.cnblogs.com/Coolxxx/p/5930385.html
