Ad Infinitum 8 - Math Programming Contest

比赛链接

A题

如果当前数是１，那么后面无论是几都会加１或者当后面数是１的时候加２，所以记录一下后面的数中１的个数（加２）即可。

如果当前数是２，那么只有当后面的数为１或者为２时才可以加１，所以再记录一下后面数中２的个数即可。

 #include <map>
 #include <cmath>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 int a[], t, n, num1[], num2[];

 int main() {
     ios::sync_with_stdio(false);
     cin >> t;
     while (t--) {
         cin >> n;
         for (int i = ; i <= n; i++) cin >> a[i];
         memset(num1, , sizeof(num1));
         memset(num2, , sizeof(num2));
         for (int i = n; i >= ; i--) {
             num1[i] = num1[i + ];
             num2[i] = num2[i + ];
             if (a[i] == ) num1[i]++;
             else if (a[i] == ) num2[i]++;
         }
         long long res = ;
         for (int i = ; i <= n; i++) {
             if (a[i] == ) {
                 res += n - i + num1[i + ];
             } else if (a[i] == ) {
                 res += num1[i + ] + num2[i + ];
             } else {
                 res += num1[i + ];
             }
         }
         cout << res << endl;
     }
     return ;
 }

B题

这题需要敏锐的洞察力啊，起码对我来说是这样！

手动向后推两项就可以发现，当Ｎ为２时，Ｓ = (a1 + 1) * (a2 + 1) - 1;

当N为３时，S = (a1 + 1) * (a2 + 1) * (a3 + 1) - 1

可以发现，一个序列的S值和其顺序无关，只和数有关。

 #include <cmath>
 #include <cstdio>
 #include <vector>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int MOD = 1e9 + ;
 const int MAX_N = ;
 const int MAX_V = ;
 typedef long long LL;
 #define rep(i, n) for (int i = (1); i <= (n); i++)

 int main() {
     ios::sync_with_stdio(false);
     int N;
     LL res = ;
     cin >> N;
     rep (i, N) {
         int x;
         cin >> x;
         res = (res + x + res * x) % MOD;
     }
     cout << res << endl;
     return ;
 }

C题

很裸的求逆，快速幂。

 #include <cmath>
 #include <cstdio>
 #include <vector>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 int extgcd(int a, int b, int &x, int &y) {
     int d = a;
     if (b != ) {
         d = extgcd(b, a % b, y, x);
         y -= (a / b) * x;
     } else {
         x = ; y = ;
     }
     return d;
 }

 int mod_inverse(int a, int m) {
     int x, y;
     extgcd(a, m, x, y);
     return (m + x % m) % m;
 }

 int mod_pow(int a, int b, int m) {
     long long res = ;
     while (b) {
         if (b & ) res = res * a % m;
         a = ((long long)a * a) % m;
         b >>= ;
     }
     return res;
 }

 int main() {
     ios::sync_with_stdio(false);
     int t;
     cin >> t;
     while (t--) {
         int a, b, x;
         cin >> a >> b >> x;
         if (b < ) {
             a = mod_inverse(a, x);
             b = -b;
         }
         cout << mod_pow(a, b, x) << endl;
     }
     return ;
 }

D题

容斥原理模板题

 #include <map>
 #include <cmath>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long LL;
 const int MAX_N = ;
 int N;
 int num[MAX_N];
 LL n, ans;
 LL gcd(LL a, LL b){
     LL r;
     while( b ){
         r = a%b;
         a = b;
         b = r;
     }
     return a;
 }

 void dfs( int id, LL Lcm, bool flag ){
     Lcm = Lcm*num[ id ] / gcd( Lcm,num[ id ] );
     if( flag ) ans += n/Lcm;
     else ans -= n/Lcm;
     for( int i=id+;i<= N;i++ ){
         dfs( i,Lcm,!flag );
     }
     return;
 }

 int main() {
     ios::sync_with_stdio(false);
     cin >> N;
     for (int i = ; i <= N; i++) cin >> num[i];
     int D;
     cin >> D;
     for (int i = ; i <= D; i++) {
         LL l, r;
         cin >> l >> r;
         LL ansr = , ansl = ;
         ans = ;
         n = r;
         for (int i = ; i <= N; i++) dfs(i, num[i], true);
         ansr = ans;
         ans = ;
         n = l - ;
         for (int i = ; i <= N; i++) dfs(i, num[i], true);
         ansl = ans;
         cout << ansr - ansl << endl;
     }
     return ;
 }

E题

从难度来看这题算是简单的规律题。。可是比赛中却没敢去写。

通过数列的前几项可以很自然的和fib数联系在一起。我们需要做的就是统计每个二进制位上的值。

通过本题学习了bitset的使用，很好用，支持两个bitset间进行位运算，很方便。

 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <vector>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int MOD = 1e9 + ;
 const int MAX_N = ;
 typedef long long LL;
 #define rep(i, n) for (int i = (1); i <= (n); i++)
 const int NUM = ;
 LL fib[NUM], sum[NUM];

 LL mod_pow(LL a, LL b) {
     LL res = ;
     while (b) {
         if (b & ) res = (res * a) % MOD;
         a = (a * a) % MOD;
         b >>= ;
     }
     return res;
 }

 bitset<NUM> solve(LL x) {
     bitset<NUM> res;
     while (x) {
         int pos = upper_bound(sum, sum + NUM, x) - sum - ;
         x -= sum[pos];
         res.set(pos);
     }
     return res;
 }

 int main() {
     fib[] = ; fib[] = ;
     sum[] = ; sum[] = ;
     for (int i = ; i < NUM; i++) {
         fib[i] = fib[i - ] + fib[i - ];
         sum[i] = sum[i - ] + fib[i];
     }

     ios_base::sync_with_stdio(false);
     int N;
     cin >> N;
     bitset<NUM> res;
     rep (i, N) {
         LL x;
         cin >> x;
         res ^= solve(x);
     }
     LL ans = ;
     for (int i = ; i < res.size(); i++) if (res.test(i)) ans = (ans + mod_pow(, i)) % MOD;
     cout << ans << endl;
     return ;
 }

F题

待续。。

G题

待续。。

H题

待续。。