1 条题解
-
1
#include <bits/stdc++.h> using namespace std; #define ls u << 1 #define rs u << 1 | 1 #define LL long long #define int long long #define PII pair <int, int> #define fi first #define se second #define pub push_back #define pob pop_back #define puf push_front #define pof pop_front #define lb lower_bound #define ub upper_bound #define i128 __int128 #define pcnt(x) __builtin_popcount(x) #define mem(a,goal) memset(a, (goal), sizeof(a)) #define rep(x,start,end) for(int x = (start) - ((start) > (end)); x != (end) - ((start) > (end)); ((start) < (end) ? x ++ : x --)) #define aLL(x) (x).begin(), (x).end() #define sz(x) (int)(x).size() const int INF = 998244353; const int P = 1e9 + 7; const int N = 2e5 + 5; int s1[N + 1][30], s2[N + 1][30]; void solve() { int n, q; cin >> n >> q; vector<char> a(n + 1), b(n + 1); for(int i = 1; i <= n; ++ i) cin >> a[i]; for(int i = 1; i <= n; ++ i) cin >> b[i]; for(int i = 1; i <= n; ++ i) { s1[i][a[i] - 'a'] ++; for(int j = 0; j < 26; ++ j) { s1[i][j] += s1[i - 1][j]; } } for(int i = 1; i <= n; ++ i) { s2[i][b[i] - 'a'] ++; for(int j = 0; j < 26; ++ j) { s2[i][j] += s2[i - 1][j]; } } while(q --) { int l, r; cin >> l >> r; int sum = 0; for(int i = 0; i < 26; ++ i) { int p = s1[r][i] - s1[l - 1][i]; int q = s2[r][i] - s2[l - 1][i]; sum += abs (p - q); } cout << sum / 2 << '\n'; } } signed main() { int t; cin >> t; while(t --) solve(); return 0; }
- 1
信息
- ID
- 5475
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 10
- 标签
- (无)
- 递交数
- 4
- 已通过
- 1
- 上传者