母函数入门 专题

HDU 1028 Ignatius and the Princess III
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1028

整数拆分
代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 155;
typedef __int64 INT;
INT c[maxn],tmp_c[maxn];
int main(){
    int n;
    while( cin >> n ){
        for(int i = 0;i <= n;++i) c[i] = 1;
        for(int i = 2;i <= n;++i){
            for(int k = 0;k <= n;k += i){
                for(int j = 0;j + k <= n;++j) tmp_c[k + j] += c[j];
            }
            for(int i = 0;i <= n;++i){
                c[i] = tmp_c[i];
                tmp_c[i] = 0;
            }
        }
        printf("%I64d\n",c[n]);
    }

    return 0;
}

HDU 1398 Square Coins
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1398
整数拆分成平方数的和

#include <bits/stdc++.h>
using namespace std;
const int maxn = 350;
typedef __int64 INT;
INT c[maxn],tmp_c[maxn];
int main(){
    int n;
    while( cin >> n && n ){
        for(int i = 0;i <= n;++i) c[i] = 1;
        for(int i = 2;i*i <= n;++i){
            for(int k = 0;k <= n;k += i*i){
                for(int j = 0;j + k <= n;++j) tmp_c[k + j] += c[j];
            }
            for(int i = 0;i <= n;++i){
                c[i] = tmp_c[i];
                tmp_c[i] = 0;
            }
        }
        printf("%I64d\n",c[n]);
    }

    return 0;
}

HDU 1085 Holding Bin-Laden Captive!
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1085
求最小的不能组成的数

#include <bits/stdc++.h>
using namespace std;
const int maxn = 10000 + 10;

typedef unsigned long long UINT;
UINT C[maxn],tmp_C[maxn];


int main(){
    int a,b,c;
    while( cin >> a >> b >> c ){
        if(!a && !b && !c) break;
        int MAX = a,ans = -1;
        memset(C,0,sizeof(C));
        memset(tmp_C,0,sizeof(tmp_C));
        for(int i = 0;i <= a;++i) C[i] = 1;
        for(int i = 0,j = 0;j <= b;++j,i += 2){
            for(int k = 0;k <= MAX;++k) tmp_C[i + k] += C[k];
        }
        MAX = a + b * 2;
        for(int k = 0;k <= MAX;++k){
            C[k] = tmp_C[k];
            tmp_C[k] = 0;
        }
        for(int i = 0,j = 0;j <= c;++j,i += 5){
            for(int k = 0;k <= MAX;++k) tmp_C[i + k] += C[k];
        }
        MAX = MAX + c * 5;
        for(int k = 0;k <= MAX;++k){
            if(!tmp_C[k]){
                ans = k;
                break;
            }
        }
        if(ans == -1) ans = MAX + 1;
        printf("%d\n",ans);
    }
}

HDU 1171 Big Event in HDU
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1171
将N个物品分成尽量等值的两堆

#include <bits/stdc++.h>
#define sf scanf
#define pf printf
using namespace std;
const int maxn = 100000 + 10;
bool C[maxn],tmp_C[maxn];
int v[maxn],cnt[maxn];
int main(){
    int n,MAX,tmp_max,ret;
    while( scanf("%d",&n) ){
        if(n < 0) break;
        ret = 0;
        for(int i = 0;i < n;++i) sf("%d %d",&v[i],&cnt[i]),ret += v[i] * cnt[i];
        memset(C,0,sizeof(C)),memset(tmp_C,0,sizeof(tmp_C));
        C[0] = 1;
        MAX = tmp_max = 0;
        for(int i = 0;i < n;++i){
            for(int j = 0,k = 0;j <= ret / 2 && k <= cnt[i];++k,j += v[i]){
                for(int p = 0;p <= MAX && p + j <= ret / 2;++p){
                    tmp_C[p + j] = tmp_C[p + j] | C[p];
                    tmp_max = max(tmp_max,p + j);
                }
            }
            for(int j = 0;j <= tmp_max;++j){
                C[j] = tmp_C[j];
                tmp_C[j] = 0;
            }
            MAX = tmp_max;
        }
        pf("%d %d\n",ret - MAX,MAX);
    }
    return 0;

}

HDU 1709 The Balance
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1709
N个数可加可减也可不操作 输出不能组成的数

#include <bits/stdc++.h>
#define sf scanf
#define pf printf
using namespace std;
const int maxn = 100 + 5 , maxm = 10000 + 5;
bool C[maxm * 2],tmp_C[maxm * 2];
int num[maxn];
int main(){
    int n,ret,ans_rt,flag;
    while( sf("%d",&n) != EOF ){
        ret = ans_rt = 0;
        memset(C,0,sizeof C),memset(tmp_C,0,sizeof tmp_C);
        C[maxm] = 1,ret = 0;
        for(int i = 0;i < n;++i) sf("%d",&num[i]),ret += num[i];
        for(int i = 0;i < n;++i){
            for(int j = maxm - ret;j <= ret + maxm;++j) tmp_C[j] = C[j];
            for(int j = maxm - ret;j + num[i] <= ret + maxm;++j){
                tmp_C[j + num[i]] = tmp_C[j + num[i]] | C[j];
            }
            for(int j = maxm - ret + num[i];j <= ret + maxm;++j){
                tmp_C[j - num[i]] = tmp_C[j - num[i]] | C[j];
            }
            for(int j = maxm - ret;j <= ret + maxm;++j){
                C[j] = tmp_C[j];
                tmp_C[j] = 0;
            }
        }
        for(int i = maxm;i <= ret + maxm;++i){
            if(!C[i]) ans_rt++;
        }
        pf("%d\n",ans_rt);
        flag = 0;
        if(ans_rt) for(int i = maxm;i <= ret + maxm;++i){
            if(!C[i]){
                if(flag) pf(" %d",i - maxm);
                else{
                    flag = 1;
                    pf("%d",i - maxm);
                }
            }
        }
        if(ans_rt) pf("\n");
    }
    return 0;
}

HDU 2069 Coin Change
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2069
最多100 个 1,5,10,25,50问能组成N的方案数

#include <bits/stdc++.h>
#define sf scanf
#define pf printf
using namespace std;
const int maxn = 255,maxm = 100 + 5;
const int MAX_CNT = 105;
int C[MAX_CNT][maxn],tmp_C[MAX_CNT][maxn];
int a[] = {5,10,25,50};

int main(){
    memset(C,0,sizeof C);
    memset(tmp_C,0,sizeof tmp_C);
    for(int i = 0;i <= 100;++i) C[i][i] = 1;
    for(int i = 0;i < 4;++i){
        for(int CNT = 0;CNT <= 100;++CNT){
            for(int j = 0;j < maxn;++j){
                for(int c = 0,k = 0;c + CNT <= 100 && j + k < maxn;c++,k += a[i]){
                    tmp_C[CNT + c][j + k] += C[CNT][j];
                }
            }
        }
        for(int CNT = 0;CNT <= 100;++CNT){
            for(int j = 0;j < maxn;++j){
                C[CNT][j] = tmp_C[CNT][j];
                tmp_C[CNT][j] = 0;
            }
        }
    }
    for(int i = 0;i < maxn;++i){
        for(int CNT = 0;CNT < 100;++CNT) C[100][i] += C[CNT][i];
    }
    int n = -1;
    while( sf("%d",&n) != EOF ) pf("%d\n",C[100][n]);
    return 0;
}