poj 3468 A Simple Problem with Integers
题意：这道题是个标准的成段更新的线段树。
View Code
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<iostream> 4 #include<string> 5 #include<queue> 6 #include<map> 7 #include<cmath> 8 #include<stack> 9 #include<algorithm> 10 #include<functional> 11 using namespace std; 12 //1.bulid(); 13 //2.query(a,b) 14 //3.update(a,b) 15 #define lson l , m , rt << 1 16 #define rson m + 1 , r , rt << 1 | 1 17 typedef long long LL; 18 const int maxn = 100055; 19 LL sign[maxn<<2]; 20 LL sum[maxn<<2]; 21 int n; 22 //根据题意做相关修改，询问时的操作 23 LL operate(LL a,LL b){ 24 return a+b; 25 } 26 void PushUp(int rt){ 27 sum[rt]=operate(sum[rt<<1],sum[rt<<1|1]); 28 } 29 void PushDown(int rt,int m) { 30 if (sign[rt]) { 31 sign[rt<<1] += sign[rt]; 32 sign[rt<<1|1] += sign[rt]; 33 sum[rt<<1] += (m - (m >> 1)) * sign[rt]; 34 sum[rt<<1|1] += (m >> 1) * sign[rt]; 35 sign[rt] = 0; 36 } 37 } 38 void bulid(int l=1,int r=n,int rt=1){ 39 sign[rt] = 0; 40 if(l==r){// 据题意做相关修改 41 scanf("%lld",&sum[rt]);return ; 42 } 43 int m=(l+r)>>1; 44 bulid(lson); 45 bulid(rson); 46 PushUp(rt); 47 } 48 void update(int L,int R,int add,int l=1,int r=n,int rt=1){ 49 if(L<=l && r<=R){// 据题意做相关修改 50 sign[rt]+=add; 51 sum[rt]+=(LL)add*(r-l+1);return ; 52 } 53 PushDown(rt,r-l+1); 54 int m = (l + r) >> 1; 55 if (L <= m) update(L , R , add , lson); 56 if (R > m) update(L , R , add , rson); 57 PushUp(rt); 58 } 59 LL query(int L,int R,int l=1,int r=n,int rt=1) { 60 if (L <= l && r <= R) { 61 return sum[rt]; 62 } 63 PushDown(rt , r - l + 1); 64 int m = (l + r) >> 1; 65 LL ret = 0; 66 if (L <= m) ret += query(L , R , lson); 67 if (m < R) ret += query(L , R , rson); 68 return ret; 69 } 70 71 int main() 72 { 73 int m,x,y,z; 74 char op[2]; 75 while(~scanf("%d%d",&n,&m)){ 76 bulid(); 77 while(m--){ 78 scanf("%s",op); 79 if(op[0]=='Q'){ 80 scanf("%d%d",&x,&y); 81 printf("%lld\n",query(x,y)); 82 }else{ 83 scanf("%d%d%d",&x,&y,&z); 84 update(x,y,z); 85 } 86 87 } 88 } 89 90 return 0; 91 }