Description
Input
Output
Sample Input
2
1 1 5 100
3
2 0 7 7
2 2
Sample Output
16
0
Data Constraint
solution
日常先打指数暴力dfs水30分
1 void dfs(int x, int y) 2 { 3 if(x == n + 1 && y == 1) 4 { 5 int sum = 0; 6 for(R int i = 1; i <= n; i++) 7 { 8 int kkk = 1; 9 for(R int j = 1; j <= a[i]; j++) 10 { 11 kkk *= b[i][j]; 12 } 13 sum = (sum + kkk) % p; 14 } 15 if(sum == c) ans++, ans %= m ; 16 return; 17 } 18 if(y == a[x]) 19 { 20 for(R int i = 0; i < p; i++) 21 { 22 b[x][y] = i; 23 dfs(x + 1, 1); 24 b[x][y] = 0; 25 } 26 return; 27 } 28 for(R int i = 0; i < p; i++) 29 { 30 b[x][y] = i; 31 dfs(x, y + 1); 32 b[x][y] = 0; 33 } 34 return; 35 } 36 signed main() 37 { 38 freopen("congruence.in","r",stdin); 39 freopen("congruence.out","w",stdout); 40 read(t); 41 while(t--) 42 { 43 read(n); read(c); read(p); read(m ); 44 ans = 0; 45 for(R int i = 1; i <= n; i++) read(a[i]); 46 dfs(1, 1); 47 writeln(ans); 48 } 49 return 0; 50 }
当n = 1 时
推一下性质
至此已经有50分了
1 void dfs(int x, int y) 2 { 3 if(x == n + 1 && y == 1) 4 { 5 int sum = 0; 6 for(R int i = 1; i <= n; i++) 7 { 8 int kkk = 1; 9 for(R int j = 1; j <= a[i]; j++) 10 { 11 kkk *= b[i][j]; 12 } 13 sum = (sum + kkk) % p; 14 } 15 if(sum == c) ans++, ans %= m ; 16 return; 17 } 18 if(y == a[x]) 19 { 20 for(R int i = 0; i < p; i++) 21 { 22 b[x][y] = i; 23 dfs(x + 1, 1); 24 b[x][y] = 0; 25 } 26 return; 27 } 28 for(R int i = 0; i < p; i++) 29 { 30 b[x][y] = i; 31 dfs(x, y + 1); 32 b[x][y] = 0; 33 } 34 return; 35 } 36 int ksm (int x,int y) 37 { 38 int res = 1; 39 while(y) 40 { 41 if(y & 1) res = res * x % m ; 42 x = x * x % m ; 43 y >>= 1; 44 } 45 return res; 46 } 47 signed main() 48 { 49 freopen("congruence.in","r",stdin); 50 freopen("congruence.out","w",stdout); 51 read(t); 52 while(t--) 53 { 54 read(n); read(c); read(p); read(m ); 55 ans = 0; 56 for(R int i = 1; i <= n; i++) read(a[i]); 57 if(n == 1) 58 { 59 if(c != 0) writeln( ksm (p - 1, a[1] - 1)); 60 else writeln( (ksm (p, a[1]) - ksm (p - 1, a[1]) + m ) % m ); 61 continue; 62 } 63 dfs(1, 1); 64 writeln(ans); 65 } 66 return 0; 67 }
先挖坑，A了再来
————————————————————————————————————————分界线
要吐了。。
每运算一次都要取模！！！！！！
不然会炸。。。
code的思路是另一种
——转自mlg%%%
记得取模
code
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<queue> 7 #include<vector> 8 #include<stack> 9 #include<set> 10 #include<deque> 11 #include<map> 12 using namespace std; 13 14 template <typename T> void read(T &x) { 15 x = 0; int f = 1; char c; 16 for (c = getchar(); c < '0' || c > '9'; c = getchar()) if (c == '-') f = -f; 17 for (; c >= '0' && c <= '9'; c = getchar()) x = 10 * x + c - '0' ; 18 x *= f; 19 } 20 template <typename T> void write(T x){ 21 if (x < 0) putchar('-'), x = -x; 22 if (x > 9) write(x / 10); 23 putchar(x % 10 + '0'); 24 } 25 template <typename T> void writeln(T x) { write(x); putchar(' '); } 26 template <typename T> void writesn(T x) { write(x); putchar(' '); } 27 28 #define int long long 29 #define inf 1234567890 30 #define next net 31 #define P 1000000007 32 #define N 50 33 #define mid ((l+r)>>1) 34 #define lson (o<<1) 35 #define rson (o<<1|1) 36 #define R register 37 38 int t, n, c, p, m ; 39 int a; 40 int a1,a2,b1,b2,ansa,ansb; 41 int ksm (int x,int y) 42 { 43 int res = 1; 44 while(y) 45 { 46 if(y & 1) res = res * x % m ; 47 x = x * x % m ; 48 y >>= 1; 49 } 50 return res; 51 } 52 inline int modd(int x) 53 { 54 x = (x % m + m ) % m ; 55 return x; 56 } 57 signed main() 58 { 59 //freopen("congruence.in","r",stdin); 60 //freopen("congruence.out","w",stdout); 61 read(t); 62 while(t--) 63 { 64 read(n); read(c); read(p); read(m ); 65 read(a); 66 ansa = a1 = a2 = modd(ksm(p - 1, a - 1)); 67 ansb = b1 = b2 = modd(ksm(p, a) - ksm(p - 1, a)); 68 for(R int i = 1; i < n; i++) 69 { 70 read(a); 71 a1 = modd(ksm(p - 1, a - 1)); 72 b1 = modd(ksm(p, a) - ksm(p - 1, a)); 73 a2 = ansa; b2 = ansb; 74 ansa = modd(modd(b1 * a2) + modd(b2 * a1) + modd(modd((p - 2) * a1) * a2)); 75 ansb = modd(modd(b1 * b2) + modd(modd((p - 1) * a1) * a2)); 76 } 77 if(c == 0) writeln(ansb); 78 else writeln(ansa); 79 } 80 return 0; 81 }