DP+高精度 URAL 1036 Lucky Tickets

题目传送门

 /*
     题意：转换就是求n位数字，总和为s/2的方案数
     DP+高精度：状态转移方程：dp[cur^1][k+j] = dp[cur^1][k+j] + dp[cur][k];
                 高精度直接拿JayYe的:)
     异或运算的规则:
     0⊕0=0,0⊕1=1
     1⊕0=1,1⊕1=0
     口诀：相同取0，相异取1
 */
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int numlen = ;
 const int numbit = ;
 const int addbit = ;
 const int maxn = numlen/numbit + ;

 struct bign {
     int len, s[numlen];
     bign() {
         memset(s, , sizeof(s));
         len = ;
     }
     bign(int num) { *this = num; }
     bign(const char *num) { *this = num; }
     bign operator = (const int num) {
         char s[numlen];
         sprintf(s, "%d", num);
         *this = s;
         return *this;
     }
     bign operator = (const char *num){
         int clen = strlen(num);
         while(clen >  && num[] == '')    num++, clen--;
         len = ;
         for(int i = clen-;i >= ; i -= numbit) {
             int top = min(numbit, i+), mul = ;
             s[len] = ;
             for(int j = ;j < top; j++) {
                 s[len] += (num[i-j]-'')*mul;
                 mul *= ;
             }
             len++;
         }
         deal();
         return *this;
     }
     void deal() {
         while(len >  && !s[len-]) len--;
     }
     bign operator + (const bign &a) const {
         bign ret;
         ret.len = ;
         int top = max(len, a.len), add = ;
         for(int i = ;add || i < top; i++) {
             int now = add;
             if(i < len) now += s[i];
             if(i < a.len)   now += a.s[i];
             ret.s[ret.len++] = now%addbit;
             add = now/addbit;
         }
         return ret;
     }
     bign operator * (const bign &a)const {
         bign ret;
         ret.len = len + a.len;
         for(int i = ;i < len; i++) {
             int pre = ;
             for(int j = ;j < a.len; j++) {
                 int now = s[i]*a.s[j] + pre;
                 pre = ;
                 ret.s[i+j] += now;
                 if(ret.s[i+j] >= addbit) {
                     pre = ret.s[i+j]/addbit;
                     ret.s[i+j] -= pre*addbit;
                 }
             }
             if(pre) ret.s[i+a.len] = pre;
         }
         ret.deal();
         return ret;
     }
 }dp[][];
 istream& operator >> (istream &in, bign &x) {
     string s;
     in >> s;
     x = s.c_str();
     return in;
 }
 ostream& operator << (ostream &out, const bign &x) {
     printf("%d", x.s[x.len-]);
     for(int i = x.len-;i >= ; i--)    printf("%04d", x.s[i]);
     return out;
 }

 int main(void)        //URAL 1036 Lucky Tickets
 {
     //freopen ("W.in", "r", stdin);

     int n, s;
     while (scanf ("%d%d", &n, &s) == )
     {
         if (s & )    {puts ("");    continue;}

         dp[][] = ;
         int cur = ;
         for (int i=; i<=n; ++i)
         {
             for (int j=; j<=s/; ++j)    dp[cur^][j] = ;
             for (int j=; j<=; ++j)
             {
                 for (int k=; k+j<=s/; ++k)
                     dp[cur^][k+j] = dp[cur^][k+j] + dp[cur][k];
             }
             cur ^= ;
         }

         cout << dp[cur][s/] * dp[cur][s/] << endl;
     }

     return ;
 }