G4int nofEvents = run->GetNumberOfEvent(); if (nofEvents == 0) return; // Merge accumulables G4AccumulableManager* accumulableManager = G4AccumulableManager::Instance(); accumulableManager->Merge(); // Compute dose = total energy deposit in a run and its variance // G4double edep = fEdep.GetValue(); G4double edep2 = fEdep2.GetValue(); G4double rms = edep2 - edep*edep/nofEvents; if (rms > 0.) rms = std::sqrt(rms); else rms = 0.; const auto detConstruction = static_cast<const DetectorConstruction*>( G4RunManager::GetRunManager()->GetUserDetectorConstruction()); G4double mass = detConstruction->GetScoringVolume()->GetMass(); G4double dose = edep/mass; G4double rmsDose = rms/mass;
时间: 2024-04-09 17:31:52 浏览: 22
这段代码是RunAction类中的一部分,用于计算模拟运行期间的剂量。
首先,代码通过调用run->GetNumberOfEvent()获取当前运行(run)中的事件数,并将其赋值给nofEvents变量。如果事件数为0,则直接返回。
接下来,代码调用G4AccumulableManager::Instance()获取累加器管理器的实例,并调用Merge()方法将累加器的值进行合并。
然后,代码分别从fEdep和fEdep2累加器中获取能量沉积的累积值和平方和的累积值,并分别赋值给edep和edep2变量。
接下来,代码计算能量沉积的均方根(RMS),通过使用公式rms = edep2 - edep*edep/nofEvents。如果rms大于0,则使用std::sqrt(rms)计算平方根,否则将rms设置为0。
接下来,代码获取用户构建的探测器构造对象(detConstruction)以及其得分体积的质量,并将其赋值给mass变量。
最后,代码计算剂量(dose)和剂量的均方根(rmsDose),分别通过将edep和rms除以mass得到。
这段代码的目的是计算模拟运行期间的能量沉积和剂量,以便后续分析和结果解释。
相关问题
用C语言改写,将打印输出的结果中重复的数去掉#include <stdio.h>#include <stdlib.h>struct ListNode { int val; struct ListNode *next;};struct ListNode* mergeList(struct ListNode* l1, struct ListNode* l2){ struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode)); // 新建一个虚拟头节点 dummy->val = 0; dummy->next = NULL; struct ListNode* cur = dummy; while (l1 && l2) { if (l1->val <= l2->val) { cur->next = l1; l1 = l1->next; } else { cur->next = l2; l2 = l2->next; } cur = cur->next; } if (l1) cur->next = l1; // 将剩余的节点加入到新链表中 if (l2) cur->next = l2; return dummy->next;}int main() { int n, m, num; scanf("%d%d", &n, &m); struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* p1 = l1; struct ListNode* p2 = l2; for (int i = 0; i < n; i++) { scanf("%d", &num); p1->next = (struct ListNode*)malloc(sizeof(struct ListNode)); p1->next->val = num; p1->next->next = NULL; p1 = p1->next; } for (int i = 0; i < m; i++) { scanf("%d", &num); p2->next = (struct ListNode*)malloc(sizeof(struct ListNode)); p2->next->val = num; p2->next->next = NULL; p2 = p2->next; } struct ListNode* res = mergeList(l1->next, l2->next); while (res) { printf("%d ", res->val); res = res->next; } return 0;}
以下是用 C 语言改写,并去掉重复数的代码:
#include <stdio.h>
#include <stdlib.h>
struct ListNode {
int val;
struct ListNode *next;
};
struct ListNode* mergeList(struct ListNode* l1, struct ListNode* l2){
struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode)); // 新建一个虚拟头节点
dummy->val = 0;
dummy->next = NULL;
struct ListNode* cur = dummy;
while (l1 && l2) {
if (l1->val <= l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
if (l1) cur->next = l1; // 将剩余的节点加入到新链表中
if (l2) cur->next = l2;
return dummy->next;
}
int main() {
int n, m, num;
scanf("%d%d", &n, &m);
struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* p1 = l1;
struct ListNode* p2 = l2;
for (int i = 0; i < n; i++) {
scanf("%d", &num);
p1->next = (struct ListNode*)malloc(sizeof(struct ListNode));
p1->next->val = num;
p1->next->next = NULL;
p1 = p1->next;
}
for (int i = 0; i < m; i++) {
scanf("%d", &num);
p2->next = (struct ListNode*)malloc(sizeof(struct ListNode));
p2->next->val = num;
p2->next->next = NULL;
p2 = p2->next;
}
struct ListNode* res = mergeList(l1->next, l2->next);
// 去除重复数
int hash[10001] = {0}; // 哈希表,假设节点值范围在 [0, 10000]
struct ListNode* cur = res;
struct ListNode* pre = NULL;
while (cur) {
if (hash[cur->val]) { // 如果当前节点值已经出现过,则删除当前节点
pre->next = cur->next;
free(cur);
cur = pre->next;
} else { // 如果当前节点值没有出现过,则将其加入哈希表,并继续遍历下一个节点
hash[cur->val] = 1;
pre = cur;
cur = cur->next;
}
}
// 输出结果
cur = res;
while (cur) {
printf("%d ", cur->val);
cur = cur->next;
}
// 释放内存
cur = res;
while (cur) {
struct ListNode* temp = cur;
cur = cur->next;
free(temp);
}
return 0;
}
Node * merge(Node*h1,Node*h2) { Node * p; p=(Node*)malloc(sizeof(Node)); while(h1!=NULL&&h2!=NULL) { if(h1->data>=h2->data) { p->next>=h2; h2=h2->next; } else{ p->next=h1; h1=h1->next; } p=p->next; } if(h1=NULL) { p->next=h2; } else p->next=h1; return p; }为什么无法将有序链表合并
这段代码中有一些错误:
1. 在第7行,应该是 `p->next = h1` 或者 `p->next = h2`,而不是 `p->next >= h2`。
2. 在第14行,应该是 `if(h1==NULL)`,而不是 `if(h1=NULL)`。
除此之外,这个函数并没有将两个有序链表合并成一个有序链表,而是创建了一个新的链表并将较小的节点插入其中,这并没有达到合并的目的。正确的合并方式应该是创建一个新的链表,依次将两个链表中较小的节点插入其中,直到一个链表为空,然后将另一个链表的剩余节点插入新链表的末尾。具体来说,可以参考下面的代码:
```
Node * merge(Node *h1, Node *h2) {
Node *p1 = h1, *p2 = h2, *newHead = NULL, *newTail = NULL;
while (p1 != NULL && p2 != NULL) {
if (p1->data <= p2->data) {
if (newHead == NULL) {
newHead = p1;
newTail = p1;
} else {
newTail->next = p1;
newTail = p1;
}
p1 = p1->next;
} else {
if (newHead == NULL) {
newHead = p2;
newTail = p2;
} else {
newTail->next = p2;
newTail = p2;
}
p2 = p2->next;
}
}
if (p1 != NULL) {
if (newHead == NULL) {
newHead = p1;
} else {
newTail->next = p1;
}
} else {
if (newHead == NULL) {
newHead = p2;
} else {
newTail->next = p2;
}
}
return newHead;
}
```