这套题目只做了几个相对简单的。其他的做起来比较吃力。
A 找下规律
1 /************************************************************************* 2 > File Name: A.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月06日 星期六 16时32分54秒 6 > Propose: 7 ************************************************************************/ 8 9 #include <cmath> 10 #include <string> 11 #include <cstdio> 12 #include <fstream> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 using namespace std; 17 /*Let's fight!!!*/ 18 19 const int MOD = 1e9 + 7; 20 typedef long long LL; 21 int t, n; 22 char s[100005]; 23 24 int main(void) { 25 ios::sync_with_stdio(false); 26 cin >> t; 27 while (t--) { 28 cin >> s; 29 n = strlen(s); 30 LL res = 1; 31 for (int i = 0; i < n; i++) { 32 if ((i + 1) % 2 == 1) res = (res * 2) % MOD; 33 else res = (res * 2 - 1) % MOD; 34 if (s[i] == 'r') res = (res + 2) % MOD; 35 } 36 cout << res << endl; 37 } 38 return 0; 39 }
B 模拟
1 n, m = map(int, raw_input().split()); 2 a = map(int, raw_input().split()); 3 now = 1; 4 while m: 5 m -= 1; 6 Q = raw_input().split(); 7 if Q[0] == 'R': 8 print a[(now + int(Q[1]) - 1) % n - 1] 9 elif Q[0] == 'C': 10 now = (now + int(Q[1])) % n; 11 else : 12 now = (now - int(Q[1]) + n) % n;
这题被我写烦了。。。是想烦了。
我是枚举中间一堆的个数。然后就是组合数学了。。。。。。。
1 /************************************************************************* 2 > File Name: C.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月06日 星期六 11时08分09秒 6 > Propose: 7 ************************************************************************/ 8 9 #include <cmath> 10 #include <string> 11 #include <cstdio> 12 #include <fstream> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 using namespace std; 17 /*Let's fight!!!*/ 18 19 const int MOD = 1e9 + 7; 20 typedef long long LL; 21 LL fact[1000005], n; 22 23 LL pow_mod(LL a, LL b) { 24 LL res = 1; 25 while (b) { 26 if (b & 1) res = (res * a) % MOD; 27 a = (a * a) % MOD; 28 b >>= 1; 29 } 30 return res; 31 } 32 33 void init() { 34 fact[5] = 1; fact[6] = 6; 35 int now = 2; 36 for (int i = 7; i <= 1000000; i++, now++) fact[i] = (fact[i - 1] * pow_mod(now, MOD - 2) % MOD * i) % MOD; 37 } 38 39 int main(void) { 40 ios::sync_with_stdio(false); 41 init(); 42 while (cin >> n) { 43 if (n < 13) { 44 cout << "0" << endl; 45 continue; 46 } 47 LL res = 0; 48 if (n % 2 == 0){ 49 for (int i = 2; i <= n - 12; i += 2) { 50 res = (res + fact[(n - i)/2 - 1]) % MOD; 51 } 52 } else { 53 for (int i = 1; i <= n - 12; i += 2) { 54 res = (res + fact[(n - i)/2 - 1]) % MOD; 55 } 56 } 57 cout << res << endl; 58 } 59 60 return 0; 61 }
D
E
题解上这题也是个Easy..可是当时愣是没有看出来，知道突破点是ｋ比较小。
正解是枚举数列中第一个和最后一个没有改变的位置, 这个不超过k^2，然后。。。
这里getchar_unlocked是getchar()的线程不安全版本。。好像已经被windows弃用。
还有，n*10可以表达为(n<<3) + (n<<1)
1 /************************************************************************* 2 > File Name: E.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月15日 星期一 21时04分57秒 6 > Propose: 7 ************************************************************************/ 8 9 #include <cmath> 10 #include <string> 11 #include <cstdio> 12 #include <fstream> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 using namespace std; 17 /*Let's fight!!!*/ 18 19 const int MAX_N = 100050; 20 typedef long long LL; 21 #define rep(i, a, b) for (int i = (a); i <= (b); i++) 22 int n, k, a[MAX_N]; 23 24 void read(int &res) { 25 char c = ' '; 26 res = 0; 27 int sign = 1; 28 while (c < '0' || c > '9') {if (c == '-') sign = -1; c = getchar_unlocked();} 29 while (c >= '0' && c <= '9') res = (res<<3) + (res<<1) + c - '0', c= getchar_unlocked(); 30 res *= sign; 31 } 32 33 int main(void) { 34 read(n), read(k); 35 rep (i, 1, n) read(a[i]); 36 37 int a1 = 0x3f3f3f3f, d = 0x3f3f3f3f; 38 rep (i, 1, k+1) rep (j, n-k+i-1, n) { 39 int cnt = 0, tmp_d = (a[j] - a[i]) / (j - i), tmp_a1 = a[i] - (i - 1) * tmp_d; 40 rep (t, 1, n) if (a[t] != tmp_a1 + (t - 1) * tmp_d) cnt++; 41 if (cnt <= k) { 42 if (a1 > tmp_a1) a1 = tmp_a1, d = tmp_d; 43 else if (a1 == tmp_a1 && d > tmp_d) d = tmp_d; 44 } 45 } 46 47 rep (i, 1, n) printf("%d%c", a1 + (i - 1) * d, i == n ? ' ' : ' '); 48 49 return 0; 50 }
1 /************************************************************************* 2 > File Name: F.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月08日 星期一 13时07分43秒 6 > Propose: 7 ************************************************************************/ 8 #include <cmath> 9 #include <string> 10 #include <cstdio> 11 #include <vector> 12 #include <fstream> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 using namespace std; 17 /*Let's fight!!!*/ 18 19 typedef long long LL; 20 LL N, M; 21 22 /* 23 * (a / b) % c 等价于 (a % (b * c)) / b; 24 * 适用于ｂｃ不大而且ｂ在ｃ的剩余系中逆元不存在的情况 25 */ 26 LL sum(LL n) { 27 LL m = M * 30; 28 n %= m; 29 return (n * (6 * n % m * n % m * n % m * n % m + 15 * n % m * n % m * n % m + 10 * n % m * n % m - 1 + m) % m) % m / 30; 30 } 31 32 int main(void) { 33 ios::sync_with_stdio(false); 34 int T; 35 cin >> T; 36 while (T--) { 37 cin >> N >> M; 38 if (N == 1) { 39 cout << N % M << endl; 40 continue; 41 } else if (N == 2) { 42 cout << (N + 16) % M << endl; 43 continue; 44 } else if (N == 3) { 45 cout << (N + 16 + 81) % M << endl; 46 continue; 47 } 48 49 LL res = 0, i = 1, fst = 0, last = N, j; 50 while (true) { 51 LL r = N / (i + 1), l = i; 52 j = N / i; 53 res = (res + (sum(last) - sum(r) + M) % M * (i % M) % M) % M; 54 if (r != fst || last != l) 55 res = (res + (sum(l) - sum(fst) + M) % M * (j % M) % M) % M; 56 if (l >= r) break; 57 i++; 58 last = r; 59 fst = l; 60 } 61 62 cout << res << endl; 63 } 64 65 return 0; 66 }
需要维护每个区间内比右短点小的个数和比左端点小的个数，BIT系列。。是个不错的题。
1 /************************************************************************* 2 > File Name: E.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月08日 星期一 17时18分23秒 6 > Propose: 7 ************************************************************************/ 8 9 #include <cmath> 10 #include <string> 11 #include <cstdio> 12 #include <fstream> 13 #include <vector> 14 #include <cstring> 15 #include <iostream> 16 #include <algorithm> 17 using namespace std; 18 /*Let's fight!!!*/ 19 20 #define rep(i, n) for (int i = (1); i <= (n); i++) 21 #define per(i, n) for (int i = (n); i >= (1); i--) 22 const int MAX_N = 200050; 23 typedef long long LL; 24 int N, M, w, A[MAX_N], c[MAX_N]; 25 struct node { 26 int id, l, r, ltr, ltl, eql, eqr; 27 node () {} 28 node (int id, int l, int r):id(id), l(l), r(r),ltr(0), ltl(0) {} 29 friend bool operator < (const node &x, const node &y) { 30 return x.id < y.id; 31 } 32 }Q[MAX_N]; 33 34 int lowbit(int x) { 35 return x & -x; 36 } 37 38 void add(int x) { 39 while (x <= w) { 40 c[x]++; 41 x += lowbit(x); 42 } 43 } 44 45 int sum(int x) { 46 int res = 0; 47 while (x > 0) { 48 res += c[x]; 49 x -= lowbit(x); 50 } 51 return res; 52 } 53 54 int main(void) { 55 ios::sync_with_stdio(false); 56 while (cin >> N >> M) { 57 vector<int> var; 58 rep (i, N) { 59 cin >> A[i]; 60 var.push_back(A[i]); 61 } 62 sort(var.begin(), var.end()); 63 var.resize(unique(var.begin(), var.end()) - var.begin()); 64 w = var.size(); 65 66 memset(c, 0, sizeof(c)); 67 LL res = 0; 68 rep (i, N) { 69 int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + 1; 70 res += i - 1 - sum(pos - 1) - (sum(pos) - sum(pos - 1)); 71 add(pos); 72 } 73 74 // cerr << res << endl; 75 76 vector<int> lft[MAX_N], rgt[MAX_N]; 77 int x, y; 78 rep (i, M) { 79 cin >> x >> y; 80 if (x > y) swap(x, y); 81 Q[i] = node(i, x, y); 82 lft[x].push_back(i); 83 rgt[y].push_back(i); 84 } 85 86 memset(c, 0, sizeof(c)); 87 rep (i, N) { 88 int szl = lft[i].size(), szr = rgt[i].size(); 89 for (int j = 0; j < szl; j++) { 90 int id = lft[i][j]; 91 int pos = lower_bound(var.begin(), var.end(), A[Q[id].r]) - var.begin() + 1; 92 Q[id].eqr = sum(pos) - sum(pos - 1); 93 Q[id].ltr = sum(pos - 1); 94 } 95 96 int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + 1; 97 for (int j = 0; j < szr; j++) { 98 int id = rgt[i][j]; 99 Q[id].ltr = sum(pos - 1) - Q[id].ltr; 100 Q[id].eqr = sum(pos) - sum(pos - 1) - Q[id].eqr; 101 } 102 add(pos); 103 } 104 105 memset(c, 0, sizeof(c)); 106 per (i, N) { 107 int szl = lft[i].size(), szr = rgt[i].size(); 108 for (int j = 0; j < szr; j++) { 109 int id = rgt[i][j]; 110 int pos = lower_bound(var.begin(), var.end(), A[Q[id].l]) - var.begin() + 1; 111 Q[id].ltl = sum(pos - 1); 112 Q[id].eql = sum(pos) - sum(pos - 1); 113 } 114 int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + 1; 115 for (int j = 0; j < szl; j++) { 116 int id = lft[i][j]; 117 Q[id].ltl = sum(pos - 1) - Q[id].ltl; 118 Q[id].eql = sum(pos) - sum(pos - 1) - Q[id].eql; 119 } 120 add(pos); 121 } 122 // cerr << "say hello "; 123 124 sort (Q + 1, Q + M + 1); 125 rep (i, M) { 126 int l = Q[i].l, r = Q[i].r; 127 LL tmp = Q[i].ltr - Q[i].ltl - (r - l + 1 - Q[i].ltr - Q[i].eqr) + (r - l + 1 - Q[i].ltl - Q[i].eql); 128 if (A[r] > A[l]) tmp--; 129 else if (A[r] < A[l]) tmp++; 130 cout << res + tmp << endl; 131 } 132 } 133 134 return 0; 135 }
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