目录
题意:。。。
析:可以直接用数状数组进行模拟,也可以用线段树。
代码如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 | #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair< int , int > P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos (-1.0); const double eps = 1e-8; const int maxn = 50000 + 5; const int mod = 1e9 + 7; const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1}; const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1}; const char *de[] = { "0000" , "0001" , "0010" , "0011" , "0100" , "0101" , "0110" , "0111" , "1000" , "1001" , "1010" , "1011" , "1100" , "1101" , "1110" , "1111" }; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min( int a, int b){ return a < b ? a : b; } inline int Max( int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in( int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int sum[maxn<<1]; int lowbit( int x){ return -x & x; } void add( int x, int val){ while (x <= n){ sum[x] += val; x += lowbit(x); } } int query( int x){ int ans = 0; while (x){ ans += sum[x]; x -= lowbit(x); } return ans; } int main(){ int T; cin >> T; for ( int kase = 1; kase <= T; ++kase){ scanf ( "%d" , &n); memset (sum, 0, sizeof sum); for ( int i = 1; i <= n; ++i){ scanf ( "%d" , &m); add(i, m); } char s[10]; int u, v; printf ( "Case %d:\n" , kase); while ( scanf ( "%s" , s) == 1 && s[0] != 'E' ){ if (s[0] == 'A' ){ scanf ( "%d %d" , &u, &v); add(u, v); } else if (s[0] == 'S' ){ scanf ( "%d %d" , &u, &v); add(u, -v); } else { scanf ( "%d %d" , &u, &v); printf ( "%d\n" , query(v)-query(u-1)); } } } return 0; } |
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