You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
InputThe first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
OutputPrint a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples input 1689 output 1869 input 18906 output 18690 ps： 比如说有一个数为1234561689，那么根据同余定理可以知道1234561689%7=（1234560000%7+1689%7）%7 然后呢，我们可以假设1689（无顺序）为后四位，那么前面所有的数%7的结果只有0~6这七种情况 所以我们可以对这七种情况打表，找出%7为0的后四位“1689”（有0的话可以让0放在最后面，其他非1689的数原样输出就好了） 代码： #include<stdio.h> #include<string.h> #define maxn 1000000+10 char a[][5]={"1869","6198","1896","1689","1986","1968","1698"}; char s[maxn]; int flag[11]={0}; int main() { scanf("%s",s); int len=strlen(s); int p=0,ps=0; for(int i=0;i<len;i++) { if((s[i]=='1'||s[i]=='6'||s[i]=='8'||s[i]=='9')&&!flag[s[i]-'0']) flag[s[i]-'0']=1; else if(s[i]=='0') ps++; else { printf("%c",s[i]); p=p*10+s[i]-'0'; p%=7; } } printf("%s",a[p]); for(int i=0;i<ps;i++) printf("0"); return 0; } 打表代码： #include<stdio.h> int a[5]={1,6,8,9},vis[20]; void dfs(int k,int s,int cnt) { if(cnt==4) { if((((s%7)+(k*10000%7))%7)==0) printf("k=%d s=%d\n",k,s); } for(int i=0;i<4;i++) { if(!vis[a[i]]) { vis[a[i]]=1; dfs(k,s*10+a[i],cnt+1); vis[a[i]]=0; } } } int main() { int i1,i2,i3,i4; for(int i=0;i<7;i++) { dfs(i,0,0); } } 总结：现在接触的数论题大多都是找规律，而且大多可以通过打表找到