POJ 3468 A Simple Problem with Integers（线段树）

题目地址：http://poj.org/problem?id=3468

思路：updata()区间增减，query()区间求和，会超int用long long

AC代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long ll;const int maxn = 100010;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;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] += (m -(m>>1)) * add[rt];        sum[rt<<1|1] += (m>>1) * add[rt];        add[rt] = 0;    }}void build(int l,int r,int rt){    add[rt] = 0;    if(l == r)    {        scanf("%I64d",&sum[rt]);        return;    }    int m = (l + r) >> 1;    build(lson);    build(rson);    pushup(rt);}void updata(int L,int R,int c,int l,int r,int rt){    if(l >= L && r <= R)    {        sum[rt] += (r-l+1) * c;        add[rt] += c;        return;    }    pushdown(rt,r-l+1);//查询或者更改的时候传递增量    int m = (l + r) >> 1;    if(L <= m) updata(L,R,c,lson);    if(R > m) updata(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);//查询或者更改的时候传递增量    int m = (l + r) >> 1;    ll ans = 0;    if(L <= m) ans += query(L,R,lson);    if(R > m) ans += query(L,R,rson);    return ans;}int main(){    int n,q;    while(scanf("%d%d",&n,&q) != EOF)    {        build(1,n,1);        char op[2];        for(int i=0; i<q; i++)        {            int a,b,c;            scanf("%s",op);            if(op[0] == 'C')            {                scanf("%d%d%d",&a,&b,&c);                updata(a,b,c,1,n,1);            }            else            {                scanf("%d%d",&a,&b);                printf("%I64d\n",query(a,b,1,n,1));            }        }    }    return 0;}