Ural 1036 Lucky Tickets

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long LL;
  4 #define MAXN 100
  5 struct HP {
  6     int len,s[MAXN];
  7     HP() {
  8         memset(s,0,sizeof(s));
  9         len = 1;
 10     }
 11     HP operator=(const char *num) { //字符串赋值
 12         len = strlen(num);
 13         for(int i = 0; i < len; i++) s[i] = num[len-i-1]-‘0‘;
 14     }
 15     HP operator=(int num) { //int 赋值
 16         char s[MAXN];
 17         sprintf(s,"%d",num);
 18         *this = s;
 19         return *this;
 20     }
 21     HP(int num) {
 22         *this = num;
 23     }
 24     HP(const char*num) {
 25         *this = num;
 26     }
 27     string str()const { //转化成string
 28         string res = "";
 29         for(int i = 0; i < len; i++) res = (char)(s[i]+‘0‘) + res;
 30         if(res == "") res = "0";
 31         return res;
 32     }
 33     HP operator +(const HP& b) const {
 34         HP c;
 35         c.len = 0;
 36         for(int i = 0,g = 0; g||i < max(len,b.len); i++) {
 37             int x = g;
 38             if(i < len) x += s[i];
 39             if(i < b.len) x += b.s[i];
 40             c.s[c.len++] = x%10;
 41             g = x/10;
 42         }
 43         return c;
 44     }
 45     void clean() {
 46         while(len > 1 && !s[len-1]) len--;
 47     }
 48 
 49     HP operator *(const HP& b) {
 50         HP c;
 51         c.len = len + b.len;
 52         for(int i = 0; i < len; i++)
 53             for(int j = 0; j < b.len; j++)
 54                 c.s[i+j] += s[i]*b.s[j];
 55         for(int i = 0; i < c.len-1; i++) {
 56             c.s[i+1] += c.s[i]/10;
 57             c.s[i] %= 10;
 58         }
 59         c.clean();
 60         return c;
 61     }
 62     HP operator - (const HP& b) {
 63         HP c;
 64         c.len = 0;
 65         for(int i = 0,g = 0; i < len; i++) {
 66             int x = s[i]-g;
 67             if(i < b.len) x -= b.s[i];
 68             if(x >= 0) g = 0;
 69             else {
 70                 g = 1;
 71                 x += 10;
 72             }
 73             c.s[c.len++]=x;
 74         }
 75         c.clean();
 76         return c;
 77     }
 78     HP operator /(const HP &b) {
 79         HP c, f = 0;
 80         for(int i = len-1; i >= 0; i--) {
 81             f = f*10;
 82             f.s[0] = s[i];
 83             while(f >= b) {
 84                 f = f - b;
 85                 c.s[i]++;
 86             }
 87         }
 88         c.len = len;
 89         c.clean();
 90         return c;
 91     }
 92     HP operator % (const HP &b) {
 93         HP r = *this / b;
 94         r = *this - r*b;
 95         return r;
 96     }
 97     HP operator /= (const HP &b) {
 98         *this  = *this / b;
 99         return *this;
100     }
101     HP operator %= (const HP &b) {
102         *this = *this % b;
103         return *this;
104     }
105     bool operator < (const HP& b) const {
106         if(len != b.len) return len < b.len;
107         for(int i = len-1; i >= 0; i--)
108             if(s[i] != b.s[i]) return s[i] < b.s[i];
109         return false;
110     }
111     bool operator > (const HP& b) const {
112         return b < *this;
113     }
114     bool operator <= (const HP& b) {
115         return !(b < *this);
116     }
117     bool operator == (const HP& b) {
118         return !(b < *this) && !(*this < b);
119     }
120     bool operator != (const HP &b) {
121         return !(*this == b);
122     }
123     HP operator += (const HP& b) {
124         *this = *this + b;
125         return *this;
126     }
127     bool operator >= (const HP &b) {
128         return *this > b || *this == b;
129     }
130 };
131 istream& operator >>(istream &in, HP& x) {
132     string s;
133     in >> s;
134     x = s.c_str();
135     return in;
136 }
137 ostream& operator <<(ostream &out, const HP& x) {
138     out << x.str();
139     return out;
140 }
141 HP dp[101][1001];
142 int main(){
143     dp[0][0] = 1;
144     for(int i = 1; i < 51; ++i)
145         for(int j = 0; j <= i*9; ++j)
146             for(int k = 0; k < 10; ++k)
147                 dp[i][j + k] = dp[i][j + k] + dp[i-1][j];
148     int n,s;
149     while(~scanf("%d%d",&n,&s)){
150         if(s&1) puts("0");
151         else cout<<dp[n][s>>1]*dp[n][s>>1]<<endl;
152     }
153     return 0;
154 }