【HDOJ】4956 Poor Hanamichi

基本数学题一道，看错位数，当成大数减做了，而且还把方向看反了。所求为最接近l的值。

 #include <cstdio>

 int f(__int64 x) {
     int i, sum;

     i = sum = ;
     while (x) {
         if (i & )
             sum -= x%;
         else
             sum += x%;
         ++i;
         x/=;
     }
     return sum;
 }

 int main() {
     __int64 l, r, x;
     bool flg;
     int t;

     scanf("%d", &t);
     while (t--) {
         scanf("%I64d%I64d", &l, &r);
         for (x=l;;++x)
             if (x% == )
                 break;
         flg = false;
         while (x <= r) {
             if (f(x) != ) {
                 flg = true;
                 break;
             }
             x += ;
         }
         if (flg)
             printf("%I64d\n", x);
         else
             printf("-1\n");
     }

     return ;
 }

好歹写出来大数减了，贴上吧。

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <string>
 using namespace std;

 string SRCB="";
 bool stop;

 int check(string s) {
     int sum=;
     int len = s.length();
     int i = , j;

     while (s[i]) {
         --len;
         j = s[i] - '';
         if (len & )
             sum -= j;
         else
             sum += j;
         ++i;
     }
     return sum;
 }

 string Minus(string a,string b) {
     string c="",ans="",t;
     int  flag=, k=;

     if(a.length()<b.length()||(a.length()==b.length()&& a.compare(b)<)) {
         t=a;
         a=b;
         b=t;
         stop=true;
         return "";
     }
     int i=a.length()-,j=b.length()-;
     while(i>=&&j>=) {
         if(a[i]+flag>b[j]) {

           c+=a[i]+flag-b[j]+'';
           flag=;
         } else if (a[i]+flag==b[j]) {
             c+='';
             flag=;
         } else {
             c+=(a[i]-'')+flag+-(b[j]-'')+'';
             flag=-;
         }
         i--;
         j--;
         k++;
     }
     while(i>=) {
         if(a[i]+flag<'') {
             c+=a[i]+flag+;
             flag=-;
         } else {
             c+=a[i]+flag;
             flag=;
         }
         i--,k++;
     }
     int len=k-;
     while(c[len]=='' && len>)
         len--;
     for(j=;j<=len;j++)
         ans+=c[j];
     char tt;
     for(i=,j=ans.length()-;i<j;i++,j--) {
         tt = ans[i];
         ans[i] = ans[j];
         ans[j]=tt;
     }
     return ans;
 }

 int main() {
     int t, tmp, ttmp;
     string l, r, ans;
     bool flag;
     char x[];
     //FILE *fout = fopen("data2", "w");

     scanf("%d", &t);
     while (t--) {
         cin >>l>>r;
         flag = stop = false;
         tmp = check(r);
         ttmp = tmp%;
         if (ttmp < )
             ttmp += ;
         if (ttmp > ) {
             x[] = ttmp-+'';
             x[] = '\0';
             ans = Minus(r, string(x));
         } else if (ttmp < ) {
             x[] = +ttmp+'';
             x[] = '\0';
             ans = Minus(r, string(x));
         } else {
             ans = r;
         }
         if (stop) {
             printf("-1\n");
             continue;
         }
         while (l.length()<ans.length() || (l.length()==ans.length()&&l.compare(ans)<=)) {
             if (check(ans) != ) {
                 flag = true;
                 break;
             }
             //cout <<ans<<endl;
             ans = Minus(ans, SRCB);
             if (stop)
                 break;
         }
         if (flag) {
             cout <<ans<<endl;
             //fprintf(fout, "%s\n", ans.data());
         } else {
             printf("-1\n");
             //fprintf(fout, "-1\n");
         }
         //fprintf(fout, "%d\n", ans);
     }
     //fclose(fout);
     return ;
 }