【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 ;

 }