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  • CodeForces 232B 组合数学 + dp

    //CodeForces 232B

    //原先 T 了的代码

     1 #include "iostream"
     2 #include "cstdio"
     3 #include "cstring"
     4 #include "algorithm"
     5 using namespace std;
     6 const int mod = 1e9 + 7;
     7 __int64 dp[110][10100];
     8 __int64 c[110][110], f[110][2];
     9 __int64 n, m, k;
    10 
    11 __int64 modEXP(__int64 a,__int64 k)
    12 {
    13     __int64 res = 1;
    14     while(k) {
    15         if(k&1)
    16             res *= a, res %= mod;
    17         a *= a;
    18         a %= mod;
    19         k >>= 1;
    20     }
    21     return res;
    22 }
    23 
    24 int main()
    25 {
    26     int i, j, r;
    27     c[0][0] = 1;
    28     for(i = 1; i<=100; ++i) {
    29         c[i][0] = 1;
    30         for(j = 1; j<=100; ++j)
    31             c[i][j] = (c[i-1][j] + c[i-1][j-1]) % mod;
    32     }
    33     scanf("%I64d%I64d%I64d", &n, &m, &k);
    34     __int64 cnt, rem;
    35     cnt = m / n;
    36     rem = m % n;
    37     for(i = 0; i<=100; ++i) {
    38         f[i][0] = modEXP(c[n][i], cnt);
    39         f[i][1] = modEXP(c[n][i], cnt + 1);
    40     }
    41     dp[0][0] = 1;
    42     for(i = 1; i <= n; ++i) {
    43         for(j = 0; j <= k; ++j) {
    44             for(r = 0; r <= n; ++r) {
    45                 if(r > j)
    46                     break;
    47                 dp[i][j] += dp[i - 1][j - r] * f[r][i<=rem] % mod;
    48                 dp[i][j] %= mod;
    49             }
    50         }
    51     }
    52     printf("%I64d
    ", dp[n][k]);
    53 }

    滚去看了数据,才发现 C(n, k) == C(n, n - k) 这里有个优化,加上就过了

     1 #include "iostream"
     2 #include "cstdio"
     3 #include "cstring"
     4 #include "algorithm"
     5 #include "cmath"
     6 using namespace std;
     7 const int mod = 1e9 + 7;
     8 __int64 dp[110][10100];
     9 __int64 c[110][110], f[110][2];
    10 __int64 n, m, k;
    11 
    12 __int64 modEXP(__int64 a,__int64 k)
    13 {
    14     __int64 res = 1;
    15     while(k) {
    16         if(k&1)
    17             res *= a, res %= mod;
    18         a *= a;
    19         a %= mod;
    20         k >>= 1;
    21     }
    22     return res;
    23 }
    24 
    25 int main()
    26 {
    27     int i, j, r;
    28     c[0][0] = 1;
    29     for(i = 1; i<=100; ++i) {
    30         c[i][0] = 1;
    31         for(j = 1; j<=100; ++j)
    32             c[i][j] = (c[i-1][j] + c[i-1][j-1]) % mod;
    33     }
    34     scanf("%I64d%I64d%I64d", &n, &m, &k);
    35     if(n<sqrt(k<<1))
    36         k = n*n - k;
    37     __int64 cnt, rem;
    38     cnt = m / n;
    39     rem = m % n;
    40     for(i = 0; i<=100; ++i) {
    41         f[i][0] = modEXP(c[n][i], cnt);
    42         f[i][1] = modEXP(c[n][i], cnt + 1);
    43     }
    44     dp[0][0] = 1;
    45     for(i = 1; i <= n; ++i) {
    46         for(j = 0; j <= k; ++j) {
    47             for(r = 0; r <= n; ++r) {
    48                 if(r > j)
    49                     break;
    50                 dp[i][j] += dp[i - 1][j - r] * f[r][i<=rem] % mod;
    51                 dp[i][j] %= mod;
    52             }
    53         }
    54     }
    55     printf("%I64d
    ", dp[n][k]);
    56 }
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  • 原文地址:https://www.cnblogs.com/AC-Phoenix/p/4286832.html
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