母函数题目

hdu 1028 Ignatius and the Princess III 母函数模板题

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

题意：

整数的拆分。

母函数学习http://www.wutianqi.com/?p=596

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a , b) ((a) < (b) ? (a) : (b))
#define Max(a , b) ((a) > (b) ? (a) : (b))#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 127
using namespace std;
//freopen("din.txt","r",stdin);int c1[N],c2[N];void init(int n){int i,j,k;for (i = 0; i <= n; ++i){c1[i] = 1;//乘完后的系数c2[i] = 0;//乘的过程中的系数
    }for (i = 2; i <= n; ++i){//总共有n个括号,从第2个起每一个括号都要和前面那一个括号相乘//所以可以忽略第一个括号 for (j = 0; j <= n; ++j){//j代表最前面这个大括号的项数 for (k = 0; k + j <= n; k += i){//在大括号后面,x都是以i方递增的 c2[k + j] += c1[j];//这里就是大括号后面的括号与前面相乘的计算 
            }}for (j = 0; j <= n; ++j){c1[j] = c2[j];c2[j] = 0;}}
}
int main(){int n;init(120);while (~scanf("%d",&n)){printf("%d\n",c1[n]);}return 0;
}

hdu 1398 Square Coins

http://acm.hdu.edu.cn/showproblem.php?pid=1398

题意：

整数拆分的题目，只不过后边的括号里面的差值变成了i^2吧了。。

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a , b) ((a) < (b) ? (a) : (b))
#define Max(a , b) ((a) > (b) ? (a) : (b))#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 327
using namespace std;
//freopen("din.txt","r",stdin);int c1[N],c2[N];
int pow2[320];void init(int n){int i,j,k;for (i = 1; i <= 300; ++i){pow2[i] = i*i;}for (i = 0; i <= n; ++i){c1[i] = 1;c2[i] = 0;}for (i = 2; i <= n; ++i){for (j = 0; j <= n; ++j){for (k = 0; k + j <= n; k += pow2[i]){c2[k + j] += c1[j];}}for (j = 0; j <= n; ++j){c1[j] = c2[j];c2[j] = 0;}}
}
int main(){int n;init(300);while (~scanf("%d",&n)){if (!n) break;printf("%d\n",c1[n]);}return 0;
}

hdu 1085 Holding Bin-Laden Captive!

http://acm.hdu.edu.cn/showproblem.php?pid=1085

题意：

给出三种硬币，价值分别为1,2,5并给出他们的数量num1,num2,num5问他们不能够组合出来的的最小价值；

思路：

前边我们介绍的都是要么是每种数量为1个或者无限个，而这里给出了确定的数量。而且询问的是不能组合出来额最小值。所以有点绕，但思路还是没有改变。

我们只要在进行乘法运算时动态的增加括号里的项数即可。最后从小到大枚举一下就好。。

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a , b) ((a) < (b) ? (a) : (b))
#define Max(a , b) ((a) > (b) ? (a) : (b))#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 8007
using namespace std;
//freopen("din.txt","r",stdin);int c1[N],c2[N];int main(){int i,j,k;int num1,num2,num5;while (~scanf("%d%d%d",&num1,&num2,&num5)){if (!num1 && !num2 && !num5) break;int MAX = num1 + num2*2 + num5*5;//本身所能表示的最大数for (i = 0; i <= MAX; ++i){c1[i] = 0;c2[i] = 0;}for (i = 0; i <= num1; ++i) c1[i] = 1;//初始化第一个括号里的项//成第二个括号里的项for (j = 0; j <= num1; ++j){for (k = 0; k <= num2*2; k += 2){c2[k + j] += c1[j];}}for (j = 0; j <= num1 + num2*2; ++j){//动态的添加了c1[j] = c2[j];c2[j] = 0;}for (j = 0; j <= num1 + num2*2; ++j){for (k = 0; k <= num5*5; k += 5){c2[k + j] += c1[j];}}for (j = 0; j <= MAX; ++j){c1[j] = c2[j];c2[j] = 0;}for (i = 1; i <= MAX; ++i){if (!c1[i]) break;}printf("%d\n",i);}return 0;
}

hdu Big Event in HDU

http://acm.hdu.edu.cn/showproblem.php?pid=1171

题意：

给出N中不同的设备，分别给出他们的价值以及数量，将这些设备分为两组，尽量使这两组的价值相等。输出各自的价值，若不相等大的在前边；

思路：

和上边一样。。。不多讲；

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a , b) ((a) < (b) ? (a) : (b))
#define Max(a , b) ((a) > (b) ? (a) : (b))#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 250007
using namespace std;
//freopen("din.txt","r",stdin);int c1[N],c2[N];int num[55],val[55];int main(){// freopen("din.txt","r",stdin);int n,i,j,k;while (~scanf("%d",&n)){if (n < 0) break;int MAX = 0;for (i = 0; i < n; ++i){scanf("%d%d",&val[i],&num[i]);MAX += (val[i]*num[i]);}for (i = 0; i <= MAX; ++i){c1[i] = c2[i] = 0;}int len = val[0]*num[0];for (i = 0; i <= len; i += val[0]){c1[i] = 1;}for (i = 1; i < n; ++i){for (j = 0; j <= len; ++j){for (k = 0; k <= val[i]*num[i]; k += val[i]){c2[k + j] += c1[j];}}len += val[i]*num[i];for (j = 0; j <= len; ++j){c1[j] = c2[j];c2[j] = 0;}}int ave = MAX/2;for (i = ave; i >= 0; --i){if (c1[i]) break;}int A = i;int B = MAX - A;printf("%d %d\n",max(A,B),min(A,B));}return 0;
}