What day is that day? ZOJ
What day is that day?
It's Saturday today, what day is it after 11 + 22 + 33 + ... + NNdays?
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
There is only one line containing one integer N (1 <= N <= 1000000000).
Output
For each test case, output one string indicating the day of week.
Sample Input
2 1 2
Sample Output
Sunday Thursday
Hint
A week consists of Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.
找規律題的一種技巧
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; #define ll long long const int N = 1e6 + 7; const int M = 500; const int mod = 7; ll ksm(ll a, ll b, ll c) { ll ans = 1; a %= c; while (b > 0) { if (b&1) ans = (ans * a) % c; a = a*a%c; b>>=1; } return ans; } int sum[N]; int next_[N]; void print(int op) { if (op == 1) { puts("Monday"); } else if ( op == 2) { puts("Tuesday"); } else if (op == 3) { puts("Wednesday"); } else if (op == 4) { puts("Thursday"); } else if (op == 5) { puts("Friday"); } else if (op == 6) { puts("Saturday"); } else { puts("Sunday"); } } void get_next(int len) { int i = 0, j = -1; next_[0] = -1; while (i <= len) { if (j == -1 || sum[i] == sum[j]) { ++i, ++j; next_[i] = j; } else j = next_[j]; } } int main() { sum[0] = 5;//因為是模7所以星期六是5、星期五是4... for (int i = 1; i <= M; ++i) { sum[i] = (ksm(i, i, mod) + sum[i-1]) % mod; } //printf ("%d\n", sum[294]); // 5 //get_next(M); int t; scanf ("%d", &t); while (t--) { int n; scanf ("%d", &n); print(sum[n%294]+1); //getchar(); //暴力法找迴圈節 // for (int i = 1; i <= n; ++i) { // int ok = 1; // for (int j = 1; j <= i; ++j) { // if (sum[j] != sum[i+j]) { // ok=0; // break; // } // } // if (ok) { // printf ("%d\n", i); // break; // } // } // kmp的next找迴圈節 //for (int i =1; i <= n; ++i) { //printf ("%d\n", sum[i]); //printf ("%d\n", next_[i]); // int len = i - next_[i]; // if (len!=i && i%len == 0 ) { // printf ("%d %d\n", i, i/len); // break; // } //} // puts("over"); } return 0; }
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