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  • hdu1028 Ignatius and the Princess III && hdu2082 找单词 && poj1664 放苹果 && noj1046 正整数划分问题——整数划分

    这种问题有两种做法,DP和母函数。

    hdu1028 Ignatius and the Princess III

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1028

    DP做法:

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstdlib>
     4 #include <cstring>
     5 #include <cctype>
     6 #include <stack>
     7 #include <queue>
     8 #include <deque>
     9 #include <map>
    10 #include <set>
    11 #include <vector>
    12 #include <cmath>
    13 #include <algorithm>
    14 #define lson l, m, rt<<1
    15 #define rson m+1, r, rt<<1|1
    16 using namespace std;
    17 typedef long long int LL;
    18 const int MAXN =  0x7fffffff;
    19 const int  MINN =  -0x7fffffff;
    20 const double eps = 1e-9;
    21 const int dir[8][2] = {{0,1},{1,0},{0,-1},{-1,0},{-1,1},
    22   {1,1},{1,-1},{-1,-1}};
    23 
    24 int main(void){
    25 #ifndef ONLINE_JUDGE
    26   freopen("hdu1028.in", "r", stdin);
    27 #endif
    28   int i, j, dp[121][121], n;
    29   dp[1][1] = 1; // WA!
    30   for (i = 2; i < 121; ++i) {
    31     dp[i][1] = dp[1][i] = 1;
    32     for (j = 2; j < 121; ++j) {
    33       if (i > j) dp[i][j] = dp[i-j][j] + dp[i][j-1];
    34       else if (i == j) dp[i][j] = dp[i][j-1] + 1;
    35       else dp[i][j] = dp[i][i];
    36     }
    37   }
    38   while (~scanf("%d", &n)) {
    39     printf("%d\n", dp[n][n]);
    40   }
    41 
    42   return 0;
    43 }

    母函数做法:

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstdlib>
     4 #include <cstring>
     5 #include <cctype>
     6 #include <stack>
     7 #include <queue>
     8 #include <deque>
     9 #include <map>
    10 #include <set>
    11 #include <vector>
    12 #include <cmath>
    13 #include <algorithm>
    14 #define lson l, m, rt<<1
    15 #define rson m+1, r, rt<<1|1
    16 using namespace std;
    17 typedef long long int LL;
    18 const int MAXN =  0x7fffffff;
    19 const int  MINN =  -0x7fffffff;
    20 const double eps = 1e-9;
    21 const int dir[8][2] = {{0,1},{1,0},{0,-1},{-1,0},{-1,1},
    22   {1,1},{1,-1},{-1,-1}};
    23 
    24 int a[121], b[121];
    25 int main(void){
    26 #ifndef ONLINE_JUDGE
    27   freopen("hdu1028.in", "r", stdin);
    28 #endif
    29   int n;
    30   while (~scanf("%d", &n)) {
    31     int i, j, k;
    32     for (i = 0; i <= n; ++i) a[i] = b[i] = 0;
    33     a[0] = 1;
    34     for (i = 1; i <= n; ++i) {
    35       for (j = 0; j <= n; ++j) {
    36         for (k = 0; k*i+j <= n; ++k) {
    37           b[k*i+j] += a[j];
    38         }
    39       }
    40       for (j = 0; j <= n; ++j) {
    41         a[j] = b[j]; b[j] = 0;
    42       }
    43     }
    44     printf("%d\n", a[n]);
    45   }
    46 
    47   return 0;
    48 }

    hdu2082 找单词

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2082

    只写了母函数的。因为这个有个数限制。

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstdlib>
     4 #include <cstring>
     5 #include <cctype>
     6 #include <stack>
     7 #include <queue>
     8 #include <deque>
     9 #include <map>
    10 #include <set>
    11 #include <vector>
    12 #include <cmath>
    13 #include <algorithm>
    14 #define lson l, m, rt<<1
    15 #define rson m+1, r, rt<<1|1
    16 using namespace std;
    17 typedef long long int LL;
    18 const int MAXN =  0x7fffffff;
    19 const int  MINN =  -0x7fffffff;
    20 const double eps = 1e-9;
    21 const int dir[8][2] = {{0,1},{1,0},{0,-1},{-1,0},{-1,1},
    22   {1,1},{1,-1},{-1,-1}};
    23 
    24 int a[51], b[51];
    25 int main(void){
    26 #ifndef ONLINE_JUDGE
    27   freopen("hdu2082.in", "r", stdin);
    28 #endif
    29   int t; scanf("%d", &t);
    30   while (t--) {
    31     int i, j, k, num;
    32     for (i = 0; i <= 50; ++i) {
    33       a[i] = b[i] = 0;
    34     }
    35     a[0] = 1;
    36     for (i = 1; i <= 26; ++i) {
    37       scanf("%d", &num);
    38       for (j = 0; j <= 50; ++j) {
    39         for (k = 0; k <= num && k*i+j <= 50; ++k) {
    40           b[k*i+j] += a[j];
    41         }
    42       }
    43       for (j = 0; j <= 50; ++j) {
    44         a[j] = b[j]; b[j] = 0;
    45       }
    46     }
    47     int sum = 0;
    48     for (i = 1; i <= 50; ++i) sum += a[i];
    49     printf("%d\n", sum);
    50   }
    51 
    52   return 0;
    53 }

    poj1664 放苹果

    题目链接:http://poj.org/problem?id=1664

    这道也只写了DP的,用DP处理比较方便。

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstdlib>
     4 #include <cstring>
     5 #include <cctype>
     6 #include <stack>
     7 #include <queue>
     8 #include <deque>
     9 #include <map>
    10 #include <set>
    11 #include <vector>
    12 #include <cmath>
    13 #include <algorithm>
    14 #define lson l, m, rt<<1
    15 #define rson m+1, r, rt<<1|1
    16 using namespace std;
    17 typedef long long int LL;
    18 const int MAXN =  0x7fffffff;
    19 const int  MINN =  -0x7fffffff;
    20 const double eps = 1e-9;
    21 const int dir[8][2] = {{0,1},{1,0},{0,-1},{-1,0},{-1,1},
    22   {1,1},{1,-1},{-1,-1}};
    23 
    24 int main(void){
    25 #ifndef ONLINE_JUDGE
    26   freopen("poj1664.in", "r", stdin);
    27 #endif
    28   int t, n, m; scanf("%d", &t);
    29   int dp[11][11], i, j;
    30   dp[1][1] = 1;
    31   for (i = 2; i < 11; ++i) {
    32     dp[i][1] = dp[1][i] = 1;
    33     for (j = 2; j < 11; ++j) {
    34       if (i > j) dp[i][j] = dp[i][j-1] + dp[i-j][j];
    35       else if (i == j) dp[i][j] = dp[i][j-1] + 1;
    36       else dp[i][j] = dp[i][i];
    37     }
    38   }
    39   while (t--) {
    40     scanf("%d%d", &m, &n);
    41     printf("%d\n", dp[m][n]);
    42   }
    43 
    44   return 0;
    45 }

    noj1046 正整数划分问题

    题目链接:http://acm.nankai.edu.cn/p1046.html

    跟hdu1028一样一样的……其实就是一道题目o(╯□╰)o

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstdlib>
     4 #include <cstring>
     5 #include <cctype>
     6 #include <stack>
     7 #include <queue>
     8 #include <deque>
     9 #include <map>
    10 #include <set>
    11 #include <vector>
    12 #include <cmath>
    13 #include <algorithm>
    14 #define lson l, m, rt<<1
    15 #define rson m+1, r, rt<<1|1
    16 using namespace std;
    17 typedef long long int LL;
    18 const int MAXN =  0x7fffffff;
    19 const int  MINN =  -0x7fffffff;
    20 const double eps = 1e-9;
    21 const int dir[8][2] = {{0,1},{1,0},{0,-1},{-1,0},{-1,1},
    22   {1,1},{1,-1},{-1,-1}};
    23 
    24 int main(void){
    25 #ifndef ONLINE_JUDGE
    26   freopen("hdu1028.in", "r", stdin);
    27 #endif
    28   int i, j, dp[121][121], n;
    29   dp[1][1] = 1; // WA!
    30   for (i = 2; i < 121; ++i) {
    31     dp[i][1] = dp[1][i] = 1;
    32     for (j = 2; j < 121; ++j) {
    33       if (i > j) dp[i][j] = dp[i-j][j] + dp[i][j-1];
    34       else if (i == j) dp[i][j] = dp[i][j-1] + 1;
    35       else dp[i][j] = dp[i][i];
    36     }
    37   }
    38   while (~scanf("%d", &n)) {
    39     printf("%d\n", dp[n][n]);
    40   }
    41 
    42   return 0;
    43 }

    感觉这些基础的东西现在都不熟练……

    哈哈,附上一首超级好听的钢琴曲~

    某青推荐的……THX~

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  • 原文地址:https://www.cnblogs.com/liuxueyang/p/3074503.html
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