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https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
71 lines
1.8 KiB
C++
71 lines
1.8 KiB
C++
#include <iostream>
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#include <cstdio>
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#include <cstring>
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#include <cmath>
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#include <algorithm>
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#include <string>
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#include <cstdlib>
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#include <vector>
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#include <set>
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#include <map>
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using namespace std;
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const double eps=1e-8;
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int len;
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//大数乘法
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void subtract(char *a,char *b,char *c)
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{
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for(int i=len-1;i>=0;i--)
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{
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c[i]=a[i]-b[i]+'0';
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if(c[i]<'0')c[i]+=10,a[i-1]--;
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}
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}
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char str[2][1005],sub[50][1005];
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int main()
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{
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int i,j,k,n;
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scanf("%d",&n);getchar();
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for(int h=1;h<=n;h++)
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{
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memset(sub,'0',sizeof(sub));
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gets(str[0]);
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len=strlen(str[0]);
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int old=1,now=0;
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int cnt=0;
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for(i=0;str[now][i];)//前置0到末尾，整个数就是0
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{
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old^=1,now^=1;//滚动数组
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if(!strcmp(str[old]+i,"10"))//10=9+1
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{
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sub[cnt][0]='9';sub[cnt][1]=0;cnt++;
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sub[cnt][0]='1';sub[cnt][1]=0;cnt++;
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break;
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}
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for(j=i;j<=len-1-j+i && str[old][j]<=str[old][len-1-j+i];j++);
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strcpy(sub[cnt],str[old]);
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if(j<len-1-j+i)//对折后有小于的，就只能取较小的回文数
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{
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sub[cnt][(i+len-1)/2]--;
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for(j=(i+len-1)/2;j>=0 && sub[cnt][j]<'0';j--)sub[cnt][j]+=10,sub[cnt][j-1]--;
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}
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for(j=0;sub[cnt][j]=='0';j++);//去前导0
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for(k=j;k<=len-1-k+j;k++)sub[cnt][len-1-k+j]=sub[cnt][k];//根据前半串对折构造后半串
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sub[cnt][len]=0;
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subtract(str[old],sub[cnt],str[now]);
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for(;str[now][i]=='0';i++);
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cnt++;
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}
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printf("Case #%d:\n%d\n",h,cnt);
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for(i=0;i<cnt;i++)
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{
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char *s;
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for(s=sub[i];*s=='0';s++);
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puts(s);
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}
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}
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return 0;
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}
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ugly problem
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