题目传送门
1 /*
2 题意：转换就是求n位数字，总和为s/2的方案数
3 DP+高精度：状态转移方程：dp[cur^1][k+j] = dp[cur^1][k+j] + dp[cur][k];
4 高精度直接拿JayYe的:)
5 异或运算的规则:
6 0⊕0=0,0⊕1=1
7 1⊕0=1,1⊕1=0
8 口诀：相同取0，相异取1
9 */
10 #include <cstdio>
11 #include <cstring>
12 #include <string>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16
17 const int numlen = 1005;
18 const int numbit = 4;
19 const int addbit = 10000;
20 const int maxn = numlen/numbit + 10;
21
22 struct bign {
23 int len, s[numlen];
24 bign() {
25 memset(s, 0, sizeof(s));
26 len = 1;
27 }
28 bign(int num) { *this = num; }
29 bign(const char *num) { *this = num; }
30 bign operator = (const int num) {
31 char s[numlen];
32 sprintf(s, "%d", num);
33 *this = s;
34 return *this;
35 }
36 bign operator = (const char *num){
37 int clen = strlen(num);
38 while(clen > 1 && num[0] == '0') num++, clen--;
39 len = 0;
40 for(int i = clen-1;i >= 0; i -= numbit) {
41 int top = min(numbit, i+1), mul = 1;
42 s[len] = 0;
43 for(int j = 0;j < top; j++) {
44 s[len] += (num[i-j]-'0')*mul;
45 mul *= 10;
46 }
47 len++;
48 }
49 deal();
50 return *this;
51 }
52 void deal() {
53 while(len > 1 && !s[len-1]) len--;
54 }
55 bign operator + (const bign &a) const {
56 bign ret;
57 ret.len = 0;
58 int top = max(len, a.len), add = 0;
59 for(int i = 0;add || i < top; i++) {
60 int now = add;
61 if(i < len) now += s[i];
62 if(i < a.len) now += a.s[i];
63 ret.s[ret.len++] = now%addbit;
64 add = now/addbit;
65 }
66 return ret;
67 }
68 bign operator * (const bign &a)const {
69 bign ret;
70 ret.len = len + a.len;
71 for(int i = 0;i < len; i++) {
72 int pre = 0;
73 for(int j = 0;j < a.len; j++) {
74 int now = s[i]*a.s[j] + pre;
75 pre = 0;
76 ret.s[i+j] += now;
77 if(ret.s[i+j] >= addbit) {
78 pre = ret.s[i+j]/addbit;
79 ret.s[i+j] -= pre*addbit;
80 }
81 }
82 if(pre) ret.s[i+a.len] = pre;
83 }
84 ret.deal();
85 return ret;
86 }
87 }dp[2][1005];
88 istream& operator >> (istream &in, bign &x) {
89 string s;
90 in >> s;
91 x = s.c_str();
92 return in;
93 }
94 ostream& operator << (ostream &out, const bign &x) {
95 printf("%d", x.s[x.len-1]);
96 for(int i = x.len-2;i >= 0; i--) printf("%04d", x.s[i]);
97 return out;
98 }
99
100 int main(void) //URAL 1036 Lucky Tickets
101 {
102 //freopen ("W.in", "r", stdin);
103
104 int n, s;
105 while (scanf ("%d%d", &n, &s) == 2)
106 {
107 if (s & 1) {puts ("0"); continue;}
108
109 dp[0][0] = 1;
110 int cur = 0;
111 for (int i=1; i<=n; ++i)
112 {
113 for (int j=0; j<=s/2; ++j) dp[cur^1][j] = 0;
114 for (int j=0; j<=9; ++j)
115 {
116 for (int k=0; k+j<=s/2; ++k)
117 dp[cur^1][k+j] = dp[cur^1][k+j] + dp[cur][k];
118 }
119 cur ^= 1;
120 }
121
122 cout << dp[cur][s/2] * dp[cur][s/2] << endl;
123 }
124
125 return 0;
126 }