poj3468-A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 46017		Accepted: 13501
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample 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

The sums may exceed the range of 32-bit integers.

 

地址：http://poj.org/problem?id=3468

 

恩这道题跟上一道hdu1698 Just a Hook很类似，就是把上道的模板拿过来改下输入输出，然后把区间成段替换改成成段增减，就是原来的赋值操作全都改成+=，然后把query填进来，不要忘了把pushudown加进去。。那个。。还有要求long long。。。还有。。。。这道题不是多组样例。。。。。。。。。一组就可以。。。。。然后，就没有然后了。。。。。。。。。

依旧是先是我的代码

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <vector>
#include <queue>
#include <algorithm>
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
using namespace std;
const int maxn = 100000;
long long sum[maxn<<2];
long long col[maxn<<2];
void PushUP(int rt) {
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m)// 此处与区间替换的差别，区间增减
{
    if (col[rt]) {
        col[rt<<1] += col[rt];
        col[rt<<1|1] += col[rt];
        sum[rt<<1] += (m - (m >> 1)) * col[rt];
        sum[rt<<1|1] += (m >> 1) * col[rt];
        col[rt] = 0;
    }
}
void build(int l,int r,int rt) {
    col[rt] = 0;
    if (l == r){
        scanf("%lld",&sum[rt]);
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUP(rt);
}
void update(int L,int R,int c,int l,int r,int rt)//区间增减
 {
    if (L <= l && r <= R) {
        col[rt] +=(long long) c;
        sum[rt] +=(long long) c * (r - l + 1);
        return ;
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (R > m) update(L , R , c , rson);
    PushUP(rt);
}
long long query(int L,int R,int l,int r,int rt)//区间求和
 {
    if (L <= l && r <= R) {
        return sum[rt];
    }
    PushDown(rt , r - l + 1);/*比单点时多的句子*/
    int m = (l + r) >> 1;
    long long ret = 0;
    if (L <= m) ret += query(L , R , lson);
    if (R > m) ret += query(L , R , rson);
    return ret;
}


int main()
{
    int n,q,a,b,c;
    char ch;
    scanf("%d%d",&n,&q);
    build(1,n,1);
    while(q--){
        scanf(" %c",&ch);
        if(ch=='Q'){
            scanf("%d%d",&a,&b);
            printf("%lld\n",query(a,b,1,n,1));
        }
        else if(ch=='C'){
            scanf("%d%d%d",&a,&b,&c);
            update(a,b,c,1,n,1);
        }
    }
    return 0;
}

然后是大神的（几乎没有任何区别。。。。。。羞愧。。。。。。我什么时候能不依赖大神自己做出来一道像样的题呢）

#include <cstdio>
#include <algorithm>
using namespace std;
 
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define LL long long
const int maxn = 111111;
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt) {
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m) {
    if (add[rt]) {
        add[rt<<1] += add[rt];
        add[rt<<1|1] += add[rt];
        sum[rt<<1] += add[rt] * (m - (m >> 1));
        sum[rt<<1|1] += add[rt] * (m >> 1);
        add[rt] = 0;
    }
}
void build(int l,int r,int rt) {
    add[rt] = 0;
    if (l == r) {
        scanf("%lld",&sum[rt]);
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt) {
    if (L <= l && r <= R) {
        add[rt] += c;
        sum[rt] += (LL)c * (r - l + 1);
        return ;
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (m < R) update(L , R , c , rson);
    PushUp(rt);
}
LL query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
        return sum[rt];
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    LL ret = 0;
    if (L <= m) ret += query(L , R , lson);
    if (m < R) ret += query(L , R , rson);
    return ret;
}
int main() {
    int N , Q;
    scanf("%d%d",&N,&Q);
    build(1 , N , 1);
    while (Q --) {
        char op[2];
        int a , b , c;
        scanf("%s",op);
        if (op[0] == 'Q') {
            scanf("%d%d",&a,&b);
            printf("%lld\n",query(a , b , 1 , N , 1));
        } else {
            scanf("%d%d%d",&a,&b,&c);
            update(a , b , c , 1 , N , 1);
        }
    }
    return 0;
}