POJ 1091 跳蚤 容斥原理

分析：其实就是看能否有一组解x1，x2, x3, x4....xn+1,使得sum{xi*ai} = 1,也就是只要有任意一个集合{ai1,ai2,ai3, ...aik|gcd(ai1, ai2, ai3...aik) = 1} ,也就是要求gcd(a1, a2, a3...an+1) = 1.an+1一定为M，那么前面n个数总共M^n种方案，减去gcd不为1的，也就是减去gcd为奇数个素数的乘积的情况，加上偶数个素数的乘积的情况。

代码：

#include <cstdio>
#include <iostream>
#include <map>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#define pb push_back
#define mp make_pair
#define esp 1e-8
#define lson   l, m, rt<<1
#define rson   m+1, r, rt<<1|1
#define sz(x) ((int)((x).size()))
#define pb push_back
#define in freopen("solve_in.txt", "r", stdin);
#define out freopen("solve_out.txt", "w", stdout);

#define bug(x) printf("Line : %u >>>>>>
", (x));
#define inf 0x7f7f7f7f
using namespace std;
typedef long long LL;
typedef map<int, int> MPS;
typedef pair<int, int> PII;

const int maxn = 211;
const int B = 10000;
struct BigInt{
    int dig[maxn], len;
    BigInt(int num = 0):len(!!num){
        memset(dig, 0, sizeof dig);
        dig[0] = num;
    }
    int operator [](int x)const{
        return dig[x];
    }
    int &operator [](int x){
        return dig[x];
    }
    BigInt normalize(){
        while(len && dig[len-1] == 0)
            len--;
        return *this;
    }
    void output(){
//        cout << len << endl;

        if(len == 0) puts("0");
        else {
            printf("%d", dig[len-1]);
            for(int i = len-2; i >= 0; i--)
                printf("%04d", dig[i]);
        }
        puts("");
    }
};
BigInt operator * (BigInt a, BigInt b){
    BigInt c;
    c.len = a.len+b.len+1;
    for(int i = 0; i < a.len; i++)
    for(int j = 0, delta = 0; j < b.len+1; j++){

        delta += a[i]*b[j]+c[i+j];
        c[i+j] = delta%B;
        delta /= B;
    }
//    c.normalize().output();
    return c.normalize();
}
BigInt operator + (BigInt a, BigInt b){

    BigInt c;
    c.len = max(a.len, b.len)+1;
    for(int i = 0, delta = 0; i < c.len; i++){
        delta += a[i]+b[i];
        c[i] = delta%B;
        delta /= B;
    }
    return c.normalize();
}
BigInt operator - (BigInt a, BigInt b){
    BigInt c;
    c.len = a.len;
    for(int i = 0, delta = 0; i < c.len; i++){
        delta += a[i]-b[i];
        c[i] = delta;
        delta = 0;
        if(c[i] < 0){
            delta = -1;
            c[i] += B;
        }
    }
    return c.normalize();
}
vector<PII> arr;
int getNum(int x) {
    int ans = 0;
    for(int i = 2; i*i <= x; i++) {
        if(x%i == 0) {
            x /= i;
            ans++;
            if(x%i == 0) return 0;
        }
    }
    if(x != 1) ans++;
    return ans;
}
BigInt getPow(int x, int y){
    BigInt res = 1;
    BigInt c;
    int len = 0;
    while(x){
        c[len] = x%B;
        c.len = max(c.len, ++len);
        x /= B;
    }
    while(y){
        if(y&1) res = res*c;
        c = c*c;
        y >>= 1;
    }
    return res;
}
int main() {

//    BigInt x = getPow(2, 1);
//    x.output();

    int n, m;
    while(scanf("%d%d", &n, &m) == 2) {
        BigInt ans = getPow(m, n);
//        ans.output();
        int y = m;
        arr.clear();
        for(int i = 1; i*i <= y; i++) if(y%i == 0) {
                int tmp = getNum(i);
                if(tmp) {
                    arr.pb(PII(i, tmp));
                }
                if(y/i != i) {
                    tmp = getNum(y/i);
                    if(tmp) {
                        arr.pb(PII(y/i, tmp));
                    }
                }
            }
        for(int i = 0; i < sz(arr); i++){
            int x = arr[i].first;
            int y = arr[i].second;
            BigInt tmp = getPow(m/x, n);
            if(y&1) ans = ans - tmp;
            else ans = ans + tmp;
        }
        ans.output();
    }
    return 0;
}

另一道很类似的题目:http://www.cnblogs.com/rootial/p/4082340.html