【题目】组合数学

生成函数

HDU 1171

题意：给你ｎ种物品(n≤50)，每种物品价值为ｖ(v≤50) ,不超过m个(m≤100)，将这些物品分配给两个人，使得两人各自物品价值和尽可能接近，输出两人各自物品价值和

思路：构造ｎ个多项式，第ｉ个多项式的ｋ×ｖ[i] 位为１，其余全为０(0 ≤ k ≤ m)，乘起来，取最接近总和的一半，且系数不为０的项的幂次作为答案。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
set<int>st;
int c1[250005],c2[250005],n;
int a[55],b[55];
int main() {
    while(scanf("%d",&n)!=EOF && n>=0) {
    memset(c1,0,sizeof(c1));
    for(int i = 1; i <= n;i++) scanf("%d%d", &a[i], &b[i]);
    c1[0] = 1;
    int s = 0;
    for(int i = 1;i <= n;i++) {
        s += a[i]*b[i];
        for(int j = 0;j <= s;j++)
        for(int k = 0,kk = 0; j + k <= s && kk <= b[i]; k += a[i], kk++)
            c2[j + k] += c1[j]; 
        for(int j = 0; j <= s;j++) {
        c1[j] = c2[j];
        c2[j] = 0;
        }
    }
    int ans;
    for(ans = s/2;ans >0 && c1[ans] ==0;ans--);
    printf("%d %d
",s-ans,ans);
    }
}

其余类似的题目：

#include <bits/stdc++.h>
using namespace std;
const int N = 105;

int a[N], b[N], n, m;
int c1[N], c2[N];
int main() {
    while(scanf("%d%d", &n, &m) != EOF) {
    for(int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]);    
    memset(c1, 0, sizeof(c1));
    memset(c2, 0, sizeof(c2));
    for(int i = a[1]; i <= b[1]; i++) c1[i] = 1;
    for(int i = 2; i <= n; i++) {
        for(int j = 0; j <= m; j++) {
        for(int k = a[i];k + j <= m && k <= b[i]; k++) 
            c2[k + j] += c1[j];
        for(int j = 0; j <= m; j++){
        c1[j] = c2[j];
        c2[j] = 0;
        }
    }
    printf("%d
", c1[m]);
    }
}

#include <bits/stdc++.h>
using namespace std;

int a[5],sum[5],d[5];
int c1[10001], c2[10001];
int main() {
    while(scanf("%d%d%d", &a[1], &a[2], &a[3]) != EOF &&(a[1] || a[2] || a[3])) {
    sum[2] = a[1] + a[2] * 2;
    sum[3] = sum[2] + a[3] * 5;
    d[2] = 2,d[3] = 5;
    memset(c1, 0, sizeof(c1));
    memset(c2, 0, sizeof(c2));
    for(int i = 0;i <= a[1]; i++) c1[i] = 1, c2[i] = 0;
    for(int i = 2;i <= 3; i++) {
        for(int j = 0;j <= sum[i]; j++){
        for(int k = 0,kk = 0;k + j <= sum[i]&&kk <= a[i];k += d[i], kk++) {
            c2[j + k] += c1[j];
        }
        }
        for(int j = 0;j <= sum[i]; j++)
        c1[j] = c2[j], c2[j] = 0;
    }
    int ans = sum[3]+1;
    for(int i = 0;i <= sum[3];i++) {
        if(c1[i] == 0) {ans = i; break;}
    }
    printf("%d
", ans);
    }
}

#include <cstdio>
#include <iostream>
using namespace std;
const int M = 10001;

int c1[M], c2[M];

int main() {
    int n;
    while(scanf("%d", &n) != EOF && n) {
    for(int i = 0; i <= n; i++) c1[i] = 1, c2[i] = 0;
    for(int i = 2; i <= 17; i++) {
       if(i*i>n)break; 
        for(int j = 0; j <= n; j++) 
        for(int k = 0; k + j <= n; k += (i*i) )
            c2[j+k] += c1[j];
        for(int j = 0; j <= n; j++)
        c1[j] = c2[j], c2[j] = 0;
       // for(int j = 0; j <= n; j++) cout << c1[j] <<" ";
       // cout<<'
';
    }
    printf("%d
", c1[n]);
    }
}

##includeinclude  <iostream><iostream>
 usingusing  namespacenamespace  stdstd;
; constconst  intint M =  M = 1000110001;

;  intint c1[M], c2[M];

 c1[M], c2[ int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
    for(int i = 0; i <= n; i++) c1[i] = 1, c2[i] = 0;
    for(int i = 2; i <= n; i++) { 
        for(int j = 0; j <= n; j++) 
        for(int k = 0; k + j <= n; k += i)
            c2[j+k] += c1[j];
        for(int j = 0; j <= n; j++)
        c1[j] = c2[j], c2[j] = 0;
    }
    printf("%d
", c1[n]);
    }
}

HDU 1521

题意：ｎ种物品，每种物品a[i]个,涌中选出ｍ个，求排列数

思路：指数型生成函数裸题，用到分数类

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

struct cre {
    ll fz,fm;
    cre operator * (const cre &o) const{
    ll z = fz * o.fz;
    ll m = fm * o.fm;
    ll _ = __gcd(z, m);
    return (cre){z/_, m/_};
    }
    cre operator + (const cre &o) const{
    ll z = fz * o.fm + o.fz * fm;
    ll m = fm * o.fm;
    ll _ = __gcd(z, m);
    return (cre){z/_, m/_};
    }
};
int jc[11];
int n, m, a[11];
cre c1[101], c2[101],tmp[101];

void ycl() {
    jc[0] = jc[1] = 1;
    for(int i = 2; i <= 10; i++) jc[i] = jc[i-1]*i;
}
int main() {
    ycl();
    while(scanf("%d%d", &n, &m)!=EOF) {
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for(int i = 0; i <= a[1]; i++) c1[i] = (cre){1,jc[i]},c2[i] = cre{0,1};
    for(int i = a[1]+1; i <= m; i++) c1[i] = c2[i] = (cre){0,1};
    for(int i = 2; i <= n; i++) {
        for(int j = 0; j <= a[i]; j++) tmp[j] = (cre){1,jc[j]};
        for(int j = a[i]+1;j <= m;j++) tmp[j] = (cre){0,1};
        for(int j = 0; j <= m; j++) {
        for(int k = 0; k <= a[i] && k+j <=m; k++){
            c2[j + k] = c2[j + k] + (tmp[k] * c1[j]);
        }
        }
        for(int j = 0; j <= m; j++) {
        c1[j] = c2[j];
        c2[j] = {0,1};
        }
    }
    if(c1[m].fm < jc[m]) {
        int bs = jc[m] / c1[m].fm;
        printf("%lld
", c1[m].fz * bs);
    }
    else {
        int bs = c1[m].fm / jc[m];
        printf("%lld
", c1[m].fz / bs);
    }
    }

}

Polya计数定理

POJ2154

题意：Ｎ个珠子串成环，用Ｎ种颜色染色，旋转重合视为一种，输出不同染色方案对Ｐ取模的答案

思路：对于某一种置换——旋转x个单位，循环节数量 m = gcd(x,N)，方案数 = N ^ m，考虑将m相同的分组，

　　　gcd(x,N) == k 则　gcd(x/k,N/k) == 1，数量为phi(n/k) ,可以枚举Ｎ的因子。

　　　最后要除以置换数Ｎ，但这里的Ｐ不一定是质数，可以在求方案数的时候的幂次减少１

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 36000;
int prime[N+5], tot;
int MOD;
bool mark[N+5];
void sss() {
    for(int i = 2;i <= N;i++) {
    if(!mark[i])   prime[++tot] = i;
    for(int j = 1; j <= tot; j++) {
        if(i * prime[j] > N) break;
        mark[i * prime[j]] = 1;
        if(i % prime[j] == 0) break;
    }
    }
} 
int phi(int x) {
    int res = x;
    for(int i = 1;prime[i]*prime[i] <= x;i++) {
    if(x % prime[i] == 0){
        res = res - res / prime[i];
        while(x % prime[i] == 0) x /= prime[i];
    }
    }
    if(x > 1) res = res - res / x;
    return res % MOD;
}
int n;
int ksm(int a,int b) {
    int ans = 1;
    a %= MOD;
    while(b) {
       if(b & 1) ans = ans * a % MOD;
       a = a * a % MOD;
       b >>= 1;       
    }
    return ans;
}
void solve() {
    int ans = 0;
    for(int i = 1;i*i <= n;i++) {
    if(n % i == 0) 
    {
        ans = (ans + phi(i) * ksm(n,n/i-1)) % MOD;
        if(n / i != i) ans = (ans + phi(n/i) * ksm(n,i-1)) % MOD;
    }
    }    
    printf("%d
", ans);
}
int main() {
    sss();
    int _;scanf("%d",&_);
    while(_--) {
    scanf("%d%d", &n, &MOD);
    solve();
    }
}

UVA10601

题意：给定12根长度相等的木棒的颜色（不超过６种颜色），求在翻转相等视为同一种的情况下，能拼成多少种不同的立方体

思路：首先分析立方体的置换

　　１．以相对面的中心旋转，90°，270°(4,4,4)，共３×２种，180°(2,2,2,2,2,2) 共３种

　　２．以对棱中心旋转，180°（1,1,2,2,2,2,2) 共６种

　　３．以对角旋转120°，240°（3,3,3,3)　共４×２种

　　４．不旋转（1,1,1,1,1,1,1,1,1,1,1,1) 共１种

综上，一共２４种置换方案，对于每一种置换，计数时做一次多维背包即可

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll dp[2][13][13][13][13][13][13];
int s[7],a[8],n;
ll jc[13];

ll DP() {
    memset(dp,0,sizeof(dp));
    dp[0][0][0][0][0][0][0] = 1;
    int o = 0;
    for(int i = 1; i <= n; i++) {
    for(int s1 = 0;s1 <=s[1];s1++)
    for(int s2 = 0;s2 <=s[2];s2++)
    for(int s3 = 0;s3 <=s[3];s3++)
    for(int s4 = 0;s4 <=s[4];s4++)
    for(int s5 = 0;s5 <=s[5];s5++)
    for(int s6 = 0;s6 <=s[6];s6++){
        if(s1 + a[i] <= s[1]) dp[o^1][s1+a[i]][s2][s3][s4][s5][s6] += dp[o][s1][s2][s3][s4][s5][s6];
        if(s2 + a[i] <= s[2]) dp[o^1][s1][s2+a[i]][s3][s4][s5][s6] += dp[o][s1][s2][s3][s4][s5][s6];
        if(s3 + a[i] <= s[3]) dp[o^1][s1][s2][s3+a[i]][s4][s5][s6] += dp[o][s1][s2][s3][s4][s5][s6];
        if(s4 + a[i] <= s[4]) dp[o^1][s1][s2][s3][s4+a[i]][s5][s6] += dp[o][s1][s2][s3][s4][s5][s6];
        if(s5 + a[i] <= s[5]) dp[o^1][s1][s2][s3][s4][s5+a[i]][s6] += dp[o][s1][s2][s3][s4][s5][s6];
        if(s6 + a[i] <= s[6]) dp[o^1][s1][s2][s3][s4][s5][s6+a[i]] += dp[o][s1][s2][s3][s4][s5][s6];
    }
    o ^= 1;
    }
    return dp[o][s[1]][s[2]][s[3]][s[4]][s[5]][s[6]];
}

int main() {
    int _;scanf("%d",&_);
    while (_--) {
    memset(s, 0, sizeof(s));
    int x;
    jc[0] = 1;
    for (int i = 1;i <= 12;i++) {
        scanf("%d", &x);
        s[x]++;
        jc[i] = jc[i-1] * i;
    }
    ll ans = 0;
    n = 3;a[1] = a[2] = a[3] = 4;
    ans += ( DP() * 6 );
    n = 6;for(int i = 1;i<=n;i++) a[i] = 2;
    ans += ( DP() * 3);
    n = 4;a[1] = a[2] = a[3] = a[4] = 3;
    ans += ( DP() * 8 );
    n = 7;a[1] = a[2] = 1;a[3] = a[4] = a[5] = a[6] = a[7] = 2;
    ans += ( DP() * 6 );
    ll ans2 = jc[12];
    for(int i = 1;i<=6;i++) ans2 /= jc[s[i]];
    ans += ans2;
    ans /= 24;
    printf("%lld
",ans);
    }
    
}

XDOJ 1228

题意：用１×２的矩阵拼成宽度为２，周长为ｎ的手环，靠近手和远离手视作不同，旋转后重合视为同一种方案，问有多少种方案

思路：polya套一个棋盘覆盖的DP，在宽度为２的情况下是一个斐波那契数列，用矩阵快速幂加速，头尾相接的方案数量等于矩阵的迹，这题可以推广宽度为ｍ的情况

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9+7;

int prime[36005],tot;
bool mark[36005];
void sss() {
    for(int i = 2;i<=36000;i++) {
    if(!mark[i]) prime[++tot] = i;
    for(int j = 1;j <= tot && i*prime[j] <= 36000; j++) {
        mark[i * prime[j]] = 1;
        if(i % prime[j] == 0) break;
    }
    }
}

int phi(int x) {
    int res = x;
    for(int i = 1;prime[i] * prime[i] <= x;i++) {
    if(x % prime[i] == 0) {
        res = res - res / prime[i];
        while(x % prime[i] == 0) x/=prime[i];
    }
    }
    if(x > 1) res = res - res / x;
    return res % MOD;
}

ll ksm (ll a,int b) {
    ll res = 1;
    while(b) {
    if(b & 1) res = res * a % MOD;
    a = a * a % MOD;
    b >>=1;
    }
    return res;
}

struct JZ{
    ll a[2][2];
    JZ operator * (const JZ &o) const {
    JZ res;
    for(int i = 0;i<=1;i++) {
        for(int j = 0;j<=1;j++) {
        ll tmp = 0;
        for(int k = 0;k<=1;k++) 
            tmp = (tmp + a[i][k] * o.a[k][j])%MOD;
        res.a[i][j] = tmp;
        }
    }
    return res;
    }
    void init(int x) {
    if(x == 1){
        a[0][0] = a[0][1] = a[1][0] = 1;
        a[1][1] = 0;
        return ;
    }
    a[0][0] = a[1][1] = 1;
        a[0][1] = a[1][0] = 0;
    }
}A,res;

pair<int,int> fib(int n) {
    A.init(1);
    res.init(0);
    while(n) {
    if(n & 1) res = res * A;
    A = A * A;
    n >>= 1;
    }
    return make_pair(res.a[0][0],res.a[1][1]);

}

ll dp(int n) {
    if(n == 1 || n == 0) return 1;
    if(n == 2) return 3;
    pair<int,int> f = fib(n);
    return (f.first + f.second) % MOD;
}

int n;


void solve() {
    ll ans = 0;
    for(int i = 1;i*i <= n;i++) {
    if(n % i == 0) {
        ans = (ans + phi(i)*1LL*dp(n/i))%MOD;
        if(n/i != i) ans = (ans + phi(n/i)*1LL*dp(i))%MOD; 
    }
    }

    ans = ans * ksm(n,MOD-2) % MOD;
    if(n % 2 == 0) ans = (ans+1LL) % MOD;
    printf("%lld
",ans);
}
int main() {
    sss();
    while(scanf ("%d",&n)!=EOF) {
    solve();
    }
}

容斥计数

UVA11806

题意：在一个ｍ行ｎ列的矩形网格里放ｋ个相同的石子，问有多少种方法，每个格子最多放一个石子，所有石子要用完，并且第一行，第一列，最后一行，最后一列都得有石子

思路：容斥搞，四种情况的并集

#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e6+7; 
typedef long long ll;


ll c[405][505];
int m,n,k;

void ycl(){
    c[1][0] = c[1][1] = 1ll;
    for(int i=2;i<=400;i++) {
    c[i][0] = 1;c[i][i] = 1ll;
    for(int j = 1;j<i;j++) c[i][j] = (c[i-1][j-1] + c[i-1][j]) % MOD;
    }
}
int main() {
    ycl();
    int _,ca=0;scanf("%d",&_);
    while(_--) {
    printf("Case %d: ",++ca);
    scanf("%d%d%d",&m,&n,&k);
    int ans = 0;
    for(int i = 0;i<(1<<4);i++) {
        int x = i,o = 1;
        int mm=m,nn=n;
        if(x&1)nn--,o=-o; x>>=1;
        if(x&1)mm--,o=-o; x>>=1;
        if(x&1)nn--,o=-o; x>>=1;
        if(x&1)mm--,o=-o; x>>=1;
        ans = ((ans + o*c[nn*mm][k])%MOD+MOD)%MOD;
    }
    if(k) printf("%d
",ans);
    else puts("0");
    }
    
}