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 >>>>>>\n", (x));
 #define inf 0x7f7f7f7f
 using namespace std;
 typedef long long LL;
 typedef map<int, int> MPS;
 typedef pair<int, int> PII;

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

         if(len == ) puts("");
         else {
             printf("%d", dig[len-]);
             for(int i = len-; i >= ; i--)
                 printf("%04d", dig[i]);
         }
         puts("");
     }
 };
 BigInt operator * (BigInt a, BigInt b){
     BigInt c;
     c.len = a.len+b.len+;
     for(int i = ; i < a.len; i++)
     for(int j = , delta = ; j < b.len+; 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)+;
     for(int i = , delta = ; 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 = , delta = ; i < c.len; i++){
         delta += a[i]-b[i];
         c[i] = delta;
         delta = ;
         if(c[i] < ){
             delta = -;
             c[i] += B;
         }
     }
     return c.normalize();
 }
 vector<PII> arr;
 int getNum(int x) {
     int ans = ;
     for(int i = ; i*i <= x; i++) {
         if(x%i == ) {
             x /= i;
             ans++;
             if(x%i == ) return ;
         }
     }
     if(x != ) ans++;
     return ans;
 }
 BigInt getPow(int x, int y){
     BigInt res = ;
     BigInt c;
     int len = ;
     while(x){
         c[len] = x%B;
         c.len = max(c.len, ++len);
         x /= B;
     }
     while(y){
         if(y&) res = res*c;
         c = c*c;
         y >>= ;
     }
     return res;
 }
 int main() {

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

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

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