# 14117: 【原4117】操作数组

### 题目描述

author: cat 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/4117

# Description

1.给定一个区间，对这个区间的每一个数加上一个给定的数；

2.查询一个给定区间的数字和。

# Input Format

Q a b 是对[a,b]进行询问；

C a,b,c是在[a,b]区间每个数加上c。

# Sanple Input

``````10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
``````

# Sample Output

``````4
55
9
15
``````

# Hint

## FineArtz's solution

``````/* 操作数组 */
#include <iostream>
using namespace std;

const int MAXN = 100000;

struct Node{
int l = 0, r = 0;
long long lazy = 0, sum = 0;
};

int n, q;
long long t[MAXN + 5];
Node a[MAXN * 4 + 5];

inline void pushUp(int x){
a[x].sum = a[x * 2].sum + a[x * 2 + 1].sum;
}

inline void pushDown(int x){
if (a[x].lazy != 0){
a[x * 2].lazy += a[x].lazy;
a[x * 2].sum += a[x].lazy * (a[x * 2].r - a[x * 2].l + 1);
a[x * 2 + 1].lazy += a[x].lazy;
a[x * 2 + 1].sum += a[x].lazy * (a[x * 2 + 1].r - a[x * 2 + 1].l + 1);
a[x].lazy = 0;
}
}

void buildTree(int x, int l, int r){
a[x].l = l;
a[x].r = r;
if (l == r){
a[x].sum = t[l];
return;
}
int mid = (l + r) / 2;
buildTree(x * 2, l, mid);
buildTree(x * 2 + 1, mid + 1, r);
pushUp(x);
}

void update(int x, int l, int r, int c){
if (a[x].l >= l && a[x].r <= r){
a[x].sum += c * (a[x].r - a[x].l + 1);
a[x].lazy += c;
return;
}
pushDown(x);
int mid = (a[x].l + a[x].r) / 2;
if (l <= mid)
update(x * 2, l, r, c);
if (r > mid)
update(x * 2 + 1, l, r, c);
pushUp(x);
}

long long query(int x, int l, int r){
if (a[x].l >= l && a[x].r <= r)
return a[x].sum;
pushDown(x);
int mid = (a[x].l + a[x].r) / 2;
long long ret = 0;
if (l <= mid)
ret += query(x * 2, l, r);
if (r > mid)
ret += query(x * 2 + 1, l, r);
return ret;
}

int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> q;
for (int i = 1; i <= n; ++i)
cin >> t[i];
buildTree(1, 1, n);
while (q--){
char op;
int x, y, z;
cin >> op;
if (op == 'Q'){
cin >> x >> y;
cout << query(1, x, y) << '\n';
}
else{
cin >> x >> y >> z;
update(1, x, y, z);
}
}
return 0;
}
``````

## ligongzzz's solution

``````#include "iostream"
#include "cstdio"
using namespace std;

int numData[100009] = { 0 };
long long numAns[100009] = { 0 };

int main() {
int N, M;
scanf("%d %d", &N, &M);

for (int i = 1; i <= N; i++) {
scanf("%d", &numData[i]);
numAns[i] = numAns[i - 1] + numData[i];
}

for (; M > 0; M--) {
char op;
scanf("\n%c", &op);

if (op == 'Q') {
int a, b;
scanf("%d %d", &a, &b);
printf("%lld\n", numAns[b] - numAns[a - 1]);
}
else {
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
if (a - 1 > N - b) {
int ans = c;
for (; a < b; a++,ans+=c)
numAns[a] += ans;
for (; a <= N; a++)
numAns[a] += ans;
}
else {
int ans = c;
for (b=b-1;b>=a;b--, ans += c)
numAns[b] -= ans;
for (; b>=0; b--)
numAns[b] -= ans;
}
}
}

return 0;
}
``````

## Neight99's solution

``````#include <iostream>

using namespace std;

const int maxN = 1e5 + 100;

long long nums[maxN];

long long question(int, int);

void plusn(int, int, long long);

int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);

int n, m, tmp1, tmp2;
char order;
long long tmp3, sum;
cin >> n >> m;

for (int i = 1; i <= n; i++) {
cin >> nums[i];
}

for (int i = 0; i < m; i++) {
cin >> order;

if (order == 'Q') {
cin >> tmp1 >> tmp2;
sum = question(tmp1, tmp2);
cout << sum << '\n';
} else if (order == 'C') {
cin >> tmp1 >> tmp2 >> tmp3;
plusn(tmp1, tmp2, tmp3);
}
}

return 0;
}

long long question(int x, int y) {
long long sum = 0;

for (int i = x; i <= y; i++) {
sum += nums[i];
}

return sum;
}

void plusn(int x, int y, long long n) {
for (int i = x; i <= y; i++) {
nums[i] += n;
}

return;
}
``````

## q4x3's solution

``````/**
* 树状数组..
* 再过不了就对lowbit打表了TATTTTTTTTTTT
* 有空写个分块
**/
#include <iostream>
#include <stdio.h>

using namespace std;

long long a[100010], b[100010], c[100010];
int N, M, tmp1, tmp2, tmp3;
char tmp;

x = 0;
char ch;
bool f = 0;
while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
if (ch == '-') f = 1;
else x = ch - '0';
while(ch = getchar(), ch >= '0' && ch <= '9') x = 10 * x + ch - '0';
x = f ? -x : x;
}

int lowbit(int x) {
return x & (-x);
}

void add(long long *a, int pos, long long val) {
for(;pos <= N;pos += lowbit(pos)) {
a[pos] += val;
}
}

long long addd(long long *a, int pos) {
long long sum = 0;
for(;pos != 0;pos -= lowbit(pos)) {
sum += a[pos];
}
return sum;
}

int main() {
int x;
for(int i = 1;i <= N;++ i) {
a[i] = x;
add(b, i, a[i] - a[i - 1]);
add(c, i, (i - 1) * (a[i] - a[i - 1]));
}
for(int i  = 0;i < M;++ i) {
while (tmp = getchar(), tmp == '\n' || tmp == '\r' || tmp == ' ');
if(tmp == 'Q') {
long long s1, s2;
s1 = (tmp1 - 1) * addd(b, tmp1 - 1) - addd(c, tmp1 - 1);
printf("%lld\n", s2 - s1);
} else {
add(b, tmp2 + 1, - tmp3);
add(c, tmp1, 1LL * tmp3 * (tmp1 - 1));
add(c, tmp2 + 1, -1LL * tmp3 * tmp2);
}
}
return 0;
}
``````

## victrid's solution

``````#include <cstdio>
#include <iostream>
using namespace std;
//The lazy tag is really confusing.
//I may just follow the tutorial.
//400005
long long seq[400005]  = {0};
long long tags[400005] = {0};
struct set {
//segmenttree
//Tree access helper
int leftbound;
int rightbound;
int current;
long long& value;
long long& tag;
set left() {
return set{
leftbound, (leftbound + rightbound) / 2, current << 1, seq[current << 1], tags[current << 1]};
}
set right() {
return set{
(leftbound + rightbound) / 2 + 1, rightbound, current << 1 | 1, seq[current << 1 | 1], tags[current << 1 | 1]};
}
return (rightbound - leftbound + 1) * tag;
}
bool isleaf() { return !(rightbound - leftbound); }
int update_fathers() {
int z = current >> 1;
while (z != 0) {
seq[z] = seq[z << 1] + seq[z << 1 | 1];
z >>= 1;
}
return 0;
}
};
void buildtree(set s) {
if (s.leftbound == s.rightbound)
cin >> s.value;
else {
buildtree(s.left());
buildtree(s.right());
s.value = s.left().value + s.right().value;
}
}
void buildtree(int total) {
buildtree(set{1, total, 1, seq[1], tags[1]});
}
int tagproc(set s) {
if (!s.isleaf()) {
if (!s.left().isleaf())
s.left().tag += s.tag;
s.left().value += s.tag * (s.left().rightbound - s.left().leftbound + 1);
if (!s.right().isleaf())
s.right().tag += s.tag;
s.right().value += s.tag * (s.right().rightbound - s.right().leftbound + 1);
}
s.tag = 0;
return 0;
}
long long getsum(int oper_left, int oper_right, set s) {
if (oper_left <= s.leftbound && s.rightbound <= oper_right) {
if (s.tag != 0)
tagproc(s);
return s.value;
} else {
long long sum = 0;
int mid       = (s.leftbound + s.rightbound) / 2;
if (s.tag != 0)
tagproc(s);
if (oper_left <= mid)
sum += getsum(oper_left, oper_right, s.left());
if (oper_right > mid)
sum += getsum(oper_left, oper_right, s.right());
return sum;
}
}
long long getsum(int oper_left, int oper_right, int total) {
return getsum(oper_left, oper_right, set{1, total, 1, seq[1], tags[1]});
}
if (oper_left <= s.leftbound && s.rightbound <= oper_right) {
s.value += add * (s.rightbound - s.leftbound + 1);
s.update_fathers();
return 0;
} else {
//if tags not passed, ancestors cannot get updated correctly.
tagproc(s);
int mid = (s.leftbound + s.rightbound) / 2;
if (oper_left <= mid) {
//Although no tags,
s.value += add * (mid - oper_left + 1);
}
if (oper_right > mid) {
s.value += add * (oper_right - mid);
}
return 0;
}
}
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
buildtree(n);
char op;
for (int i = 0; i < m; i++) {
do {
op = getchar();
} while (op == ' ' || op == '\n');
if (op == 'Q') {
scanf("%d %d", &oper_left, &oper_right);
printf("%lld\n", getsum(oper_left, oper_right, n));
} else {
scanf("%d %d %d", &oper_left, &oper_right, &oper_add);
}
}
return 0;
}
``````

## WashSwang's solution

``````#include <iostream>
#include <cstdio>
using namespace std;
const int MAXN=100001;
typedef long long ll;
ll a[MAXN],ans[4*MAXN],tag[4*MAXN];
inline int ls(int p) {
return p<<1;
}

inline int rs(int p) {
return p<<1|1;
}

inline void push_up(int p) {
ans[p]=ans[ls(p)]+ans[rs(p)];
}

void build(int l,int r,int p) {
if (l==r)
{
ans[p]=a[l];
return;
}
int mid=(l+r)>>1;
build(l,mid,ls(p));
build(mid+1,r,rs(p));
push_up(p);
}

inline void add_tag(int p,int l,int r,int k)
{
tag[p]+=k;
ans[p]+=k*(r-l+1);
}

inline void push_down(int p,int l,int r)
{
int mid=(l+r)>>1;
tag[p]=0;
}

void update(int nl,int nr,int l,int r,int p,int k)
{
if (nl<=l&&r<=nr)
{
return;
}
push_down(p,l,r);
int mid=(l+r)>>1;
if (nl<=mid) update(nl,nr,l,mid,ls(p),k);
if (nr>mid) update(nl,nr,mid+1,r,rs(p),k);
push_up(p);
}

ll query(int nl,int nr,int l,int r,int p)
{
if (nl<=l&&r<=nr) return ans[p];
push_down(p,l,r);
int mid=(l+r)>>1;
ll sum=0;
if (nl<=mid) sum+=query(nl,nr,l,mid,ls(p));
if (nr>mid) sum+=query(nl,nr,mid+1,r,rs(p));
return sum;
}
int m,n,x,y,k;
char c;
int main()
{
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++) scanf("%lld",&a[i]);
build(1,n,1);
for (int i=0;i<m;++i){
c=' ';
while (c!='C'&&c!='Q') c=getchar();
if (c=='C')
{
scanf("%d%d%d",&x,&y,&k);
update(x,y,1,n,1,k);
}
if (c=='Q')
{
scanf("%d%d",&x,&y);
printf("%lld\n",query(x,y,1,n,1));
}
}
return 0;
}
``````

## yyong119's solution

``````#include <iostream>
#include <cstdio>
#define MAX_N 100010
using namespace std;

long long tree[MAX_N << 2];
int flag[MAX_N << 2];

void buildtree(int idx, int l, int r) {
if (l == r) {
scanf("%lld", &tree[idx]);
return;
}
int mid = (l + r) >> 1;
buildtree(idx << 1, l, mid);
buildtree(idx << 1 | 1, mid + 1, r);
tree[idx] = tree[idx << 1] + tree[idx << 1 | 1];
}

void pushdown(int idx, int l, int r) {
int mid = (l + r) >> 1;
tree[idx << 1] += (long long) flag[idx] * (mid - l + 1);
tree[idx << 1 | 1] += (long long) flag[idx] * (r - mid);
flag[idx << 1] += flag[idx];
flag[idx << 1 | 1] += flag[idx];
flag[idx] = 0;
}

void add(int idx, int l, int r, int a, int b, int c) {
if (l == a && b == r) {
tree[idx] += (long long)c * (b - a + 1);
flag[idx] += c;
return;
}
if (flag[idx])
pushdown(idx, l, r);
int mid = (l + r) >> 1;
if (b <= mid)
add(idx << 1, l, mid, a, b, c);
else if (a > mid)
add(idx << 1 | 1, mid + 1, r, a, b, c);
else {
add(idx << 1, l, mid, a, mid, c);
add(idx << 1 | 1, mid + 1, r, mid + 1, b, c);
}
tree[idx] = tree[idx << 1] + tree[idx << 1 | 1];
}

long long query(int idx, int l, int r, int a, int b) {
if (l == a && r == b)
return tree[idx];
int mid = (l + r) >> 1;
if (flag[idx])
pushdown(idx, l, r);
if (b <= mid)
return query(idx << 1, l, mid, a, b);
else if (a > mid)
return query(idx << 1 | 1, mid + 1, r, a, b);
else
return query(idx << 1, l, mid, a, mid) + query(idx << 1 | 1, mid + 1, r, mid + 1, b);
}

int main() {

int n, m;
scanf("%d%d", &n, &m);
buildtree(1, 1, n);
while (m--) {
char op[2]; scanf("%s", op);
if (op[0] == 'Q') {
int a, b; scanf("%d%d", &a, &b);
printf("%lld\n", query(1, 1, n, a, b));
}
else {
int a, b, c; scanf("%d%d%d", &a, &b, &c);
add(1, 1, n, a, b, c);
}
}
return 0;
}
``````