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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1166
思路分析:该问题为动态连续和查询问题,使用数组数组可以解决;也可使用线段树解决该问题;
代码如下:
#include <cstdio> #include <cstring> #include <iostream> using namespace std; const int MAX_N = 50000 + 10; int c[MAX_N]; inline int Lowbit(int x) { return x & -x; } inline int Sum(int x) { int ret = 0; while (x) { ret += c[x]; x -= Lowbit(x); } return ret; } inline void Add(int x, int d, int n) { while (x <= n) { c[x] += d; x += Lowbit(x); } } int main() { int a = 0, b = 0, ans = 0, temp = 0; int test_case = 0, case_id = 0, n = 0; char command[10]; scanf("%d", &test_case); while (test_case--) { scanf("%d", &n); memset(c, 0, sizeof(c)); for (int i = 1; i <= n; ++ i) { scanf("%d", &temp); Add(i, temp, n); } printf("Case %d:\n", ++case_id); while(scanf("%s", command) != EOF && command[0] != ‘E‘) { scanf("%d %d", &a, &b); if (command[0] == ‘Q‘) { ans = Sum(b) - Sum(a - 1); printf("%d\n", ans); } else if (command[0] == ‘A‘) { Add(a, b, n); } else { Add(a, -b, n); } } } return 0; }
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原文地址:http://www.cnblogs.com/tallisHe/p/4657499.html