Lucky Tickets
Time Limit: 2000ms
Memory Limit: 16384KB
This problem will be judged on Ural. Original ID: 103664-bit integer IO format: %lld Java class name: (Any)
You are given a number 1 ≤ N ≤ 50. Every ticket has its 2N-digit number. We call a ticket lucky, if the sum of its first N digits is equal to the sum of its last N digits. You are also given the sum of ALL digits in the number. Your task is to count an amount of lucky numbers, having the specified sum of ALL digits.
Input
Two space-separated numbers: N and S. Here S is the sum of all digits. Assume that 0 ≤ S ≤ 1000.
Output
The amount of lucky tickets.
Sample Input
2 2
Sample Output
4
Hint
The tickets are 0101, 0110, 1001, 1010 in the example above
解题：动态规划$dp[i][j]表示处理到了第i位，且前i位的和为j的方案数$
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 }