Sereja and Array-数组操作或者线段树或树状数组

CodeForces - 315B Sereja and Array Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description Sereja has got an array, consisting of n integers, a1, a2, ..., an. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms: Make vi-th array element equal to xi. In other words, perform the assignment avi = xi. Increase each array element by yi. In other words, perform n assignments ai = ai + yi(1 ≤ i ≤ n). Take a piece of paper and write out the qi-th array element. That is, the element aqi. Help Sereja, complete all his operations. Input The first line contains integers n, m(1 ≤ n, m ≤ 105). The second line contains n space-separated integers a1, a2, ..., an(1 ≤ ai ≤ 109) — the original array. Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer ti(1 ≤ ti ≤ 3) that represents the operation type. If ti = 1, then it is followed by two integers vi and xi, (1 ≤ vi ≤ n, 1 ≤ xi ≤ 109). If ti = 2, then it is followed by integer yi(1 ≤ yi ≤ 104). And if ti = 3, then it is followed by integer qi(1 ≤ qi ≤ n). Output For each third type operation print value aqi. Print the values in the order, in which the corresponding queries follow in the input. Sample Input Input 10 11 1 2 3 4 5 6 7 8 9 10 3 2 3 9 2 10 3 1 3 10 1 1 10 2 10 2 10 3 1 3 10 3 9 Output 2 9 11 20 30 40 39 这道题目大家需要思考，不要一看到题目就用线段树，要想想有没有更好的方法，这里的题目给出的三种操作 可以知道没有一个是对区间进行操作的，唯一一个都是对整个数组操作，对所有的数的影响一样。 所以代码便可以变为如下： /* Author: 2486 Memory: 204 KB Time: 93 MS Language: GNU G++11 4.9.2 Result: Accepted VJ RunId: 4206974 Real RunId: 12270208 Public: No Yes */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=1e5+5; int n,m,p,c,v,a[maxn]; int main() { scanf("%d%d",&n,&m); for(int i=1; i<=n; i++) { scanf("%d",&a[i]); } int cnt=0; while(m--) { scanf("%d%d",&p,&c); if(p==1) { scanf("%d",&v); a[c]=v-cnt; } else if(p==2) { cnt+=c; } else { printf("%d\n",a[c]+cnt); } } return 0; }