/* A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 44944 Accepted: 13169 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000. Each of the next Q lines represents an operation. "C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000. "Q a b" means querying the sum of Aa, Aa+1, ... , Ab. 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. Source POJ Monthly--2007.11.25, Yang Yi 题意:略 */ /* 本题我做的时候出了点差错,后来看了大神的才知道 */ #include <iostream> #include<algorithm> #include<queue> #include<cmath> #include<string.h> #include<stdio.h> #include<stdlib.h> 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];//不用长整型wa 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];//和乘法(赋相同的值)的区别,乘法是直接赋值 add[rt<<1|1]+=add[rt];//和乘法(赋相同的值)的区别,乘法是直接赋值 sum[rt<<1] +=(m-(m>>1))*add[rt];//长度和 (l+r)/2-l一样即(r-l+1)-(r-l+1)/2==(l+r)/2-l sum[rt<<1|1] += (m>>1)*add[rt]; add[rt] = 0; } } void cre(int l,int r,int rt) { add[rt]=0; if(l==r) { scanf("%lld",&sum[rt]); return; } int m= (l+r)>>1; cre(lson); cre(rson); PushUp(rt); } void upd(int L,int R,int c,int l,int r,int rt) { // printf("前面 L = %d R = %d l = %d r = %d rt = %d ",L,R,l,r,rt); if(L<=l&&r<=R) { add[rt]+=c;//和乘法(赋相同的值)的区别,乘法是直接赋值 sum[rt]+=(LL)c*(r-l+1); // printf("L = %d R = %d l = %d r = %d rt = %d sum[%d] = %d ",L,R,l,r,rt,rt,sum[rt]); return ; } PushDown(rt,r-l+1); int m =(l+r)>>1; if(L<=m) upd(L,R,c,lson); if(R>m) upd(L,R,c,rson); PushUp(rt); } LL qur(int L,int R,int l,int r,int rt) { // printf("L = %d R = %d l = %d r = %d rt = %d ",L,R,l,r,rt); // system("pause"); 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+=qur(L,R,lson); //printf("ret = %d ",ret); } if(R>m) { ret+=qur(L,R,rson); // printf("ret = %d ",ret); } return ret; } int main() { int N,Q,a,b,c,i; char ch[2]; while(~scanf("%d %d",&N,&Q)) { cre(1,N,1); for(i=1; i<=Q; i++) { scanf("%s",ch); if(ch[0]=='Q') { scanf("%d%d",&a,&b); printf("%lld ",qur(a,b,1,N,1)); } else if(ch[0]=='C') { scanf("%d%d%d",&a,&b,&c); upd(a,b,c,1,N,1); } } } return 0; }