$sum_{k = 0} ^ {n} a_kx^k = sum_{k = 0} ^ {n} b_k(x - t)^k Leftrightarrow sum_{k = 0} ^ {n} a_k(x + t)^k = sum_{k = 0} ^ {n} b_kx^k$
把式子左边用二项式定理展开：
$b_m=sum_{k=m}^nC_k^{k-m}t^{k-m}a_k\=sum_{k=0}^{n-m}C_{m+k}^k t^k a_{m+k}$
注意到$n - m leq 5$，然后。。。然后就没有然后了
出题人都是*****！
1 /************************************************************** 2 Problem: 3933 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:324 ms 7 Memory:1148 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 typedef long long ll; 16 const int Len = 2e3 + 5; 17 const int N = 3e3 + 5; 18 const ll base = 1e9; 19 const int mod = 3389; 20 21 inline void print(int t, bool f); 22 23 struct Big { 24 ll x[Len]; 25 int len; 26 Big(ll _ = 0) { 27 memset(x, 0, sizeof(x)); 28 x[len = 1] = _; 29 } 30 31 inline ll& operator [] (int i) { 32 return x[i]; 33 } 34 35 inline void get() { 36 static char st[N]; 37 static int i, l, tmp; 38 memset(x, 0, sizeof(x)); 39 gets(st + 1), l = strlen(st + 1); 40 for (i = l; i; --i) { 41 if ((l - i) % 9 == 0) tmp = 1; 42 x[(l - i) / 9 + 1] += 1ll * (st[i] - '0') * tmp, tmp *= 10; 43 } 44 len = (l - 1) / 9 + 1; 45 } 46 inline void put(char ch = ' ') { 47 static int i; 48 for (print(x[len], 0), i = len - 1; i; --i) 49 print(x[i], 1); 50 putchar(ch); 51 } 52 53 inline Big operator + (const Big &b) const { 54 static Big res; 55 static int i; 56 res = 0, res.len = max(len, b.len); 57 for (i = 1; i <= res.len; ++i) { 58 res[i] += x[i] + b.x[i]; 59 if (res[i] >= base) 60 res[i] -= base, ++res[i + 1]; 61 } 62 if (res[res.len + 1]) ++res.len; 63 while (!res[res.len]) --res.len; 64 return res; 65 } 66 inline Big& operator += (const Big &b) { 67 return *this = *this + b; 68 } 69 inline Big& operator ++() { 70 return *this += 1; 71 } 72 73 inline ll operator - (const Big &b) { 74 static int i; 75 static ll res; 76 for (res = 0, i = len; i; --i) 77 res = res * base + x[i] - b.x[i]; 78 return res; 79 } 80 81 inline Big operator * (const Big &b) const { 82 static Big res; 83 static int i, j; 84 res = 0, res.len = len + b.len; 85 for (i = 1; i <= len; ++i) 86 for (j = 1; j <= b.len; ++j) { 87 res[i + j - 1] += x[i] * b.x[j]; 88 res[i + j] += res[i + j - 1] / base, res[i + j - 1] %= base; 89 } 90 for (i = 1; i <= res.len; ++i) 91 res[i + 1] += res[i] / base, res[i] %= base; 92 while (!res[res.len] && res.len) --res.len; 93 if (!res.len) res.len = 1; 94 return res; 95 } 96 inline Big& operator *= (const Big &b) { 97 *this = *this * b; 98 } 99 100 inline Big operator / (int t) const { 101 static Big res; 102 static int i; 103 res = *this; 104 for (i = len; i; --i) 105 res[i - 1] += (res[i] % t) * base, res[i] /= t; 106 res[0] = 0; 107 while (!res[res.len] && res.len) --res.len; 108 if (!res.len) res.len = 1; 109 return res; 110 } 111 inline Big& operator /= (int t) { 112 return *this = *this / t; 113 } 114 115 ll operator % (int t) const { 116 static int i; 117 static ll res; 118 for (res = 0, i = len; i; --i) 119 res = (res * base + x[i]) % t; 120 return res; 121 } 122 } n, m, ans, C, p, t; 123 124 inline int pow(int x, int y) { 125 static int res; 126 res = 1; 127 while (y) { 128 if (y & 1) res = res * x % mod; 129 x = x * x % mod, y >>= 1; 130 } 131 return res; 132 } 133 134 inline int A(const Big &t) { 135 return (209 * pow(1234, t % (mod - 1)) + 3181) % mod; 136 } 137 138 int main() { 139 int i, tmp; 140 n.get(), t.get(), m.get(); 141 tmp = n - m, C = p = 1, ans = 0; 142 for (i = 0; i <= tmp; ++i, ++m) { 143 if (i) C = (C * m) / i, p *= t; 144 ans += C * p * A(m); 145 } 146 ans.put(); 147 return 0; 148 } 149 150 inline void print(int t, bool f) { 151 static int tmp, tot, pr[10]; 152 if (t < 0) putchar('-'), tmp = -t; 153 else tmp = t; 154 tot = 0; 155 while (tmp) 156 pr[++tot] = tmp % 10, tmp /= 10; 157 if (f) for (tmp = 9 - tot; tmp; --tmp) putchar('0'); 158 while (tot) putchar(pr[tot--] + '0'); 159 }