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POJ3468 A Simple Problem with Integers

2014-05-04 16:27 387 查看
PS: 线段树区间操作,省赛热身。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100100;
typedef long long LL;
int N, Q;
LL sumv[maxn*3], addv[maxn*3];
int ql, qr;
LL _sum;
//Accepted 4836K 1750MS C++ 1949B 2014-05-04 16:23:50
void maintain(int o, int L, int R) {
int lc = o*2, rc = o*2+1;
if(L < R) {
sumv[o] = sumv[lc] + sumv[rc];
}
sumv[o] += (R-L+1)*addv[o]; //WA.
if(L==R) {
addv[o] = 0;
}
}

void update(int o, int L, int R, int v) {
if(ql<=L && qr >= R) {
addv[o] += v;
} else {
int M = L + (R-L)/2;
if(ql <= M) update(o*2, L, M, v);
if(qr > M) update(o*2+1, M+1, R, v);
}
maintain(o, L, R);
}

void query(int o, int L, int R, LL add) {
if(ql<=L && qr >= R) {
_sum += sumv[o] + add*(R-L+1);
} else {
int M = L + (R-L)/2;
if(ql <= M) query(o*2, L, M, add+addv[o]);
if(qr > M) query(o*2+1, M+1, R, add+addv[o]);
}
}

void init() {
memset(sumv, 0, sizeof(sumv));
memset(addv, 0, sizeof(addv));
}

int main()
{
// freopen("in", "r", stdin);
int tmp, from, to;
char str[5];
while(scanf("%d%d", &N, &Q)!=EOF) {
init();
for(int i = 1; i <= N; i++) {
scanf("%d", &tmp);
ql = qr = i;
update(1, 1, N, tmp);
}
getchar();
for(int i = 1; i <= Q; i++) {
scanf("%s", str);
int v = 0;
if(str[0]=='Q') {
scanf("%d%d", &from, &to);
ql = from, qr = to;
_sum = 0, v = 0;
query(1, 1, N, v);
printf("%lld\n", _sum);
}
else if(str[0]=='C') {
int add;
scanf("%d%d%d", &from, &to, &add);
ql = from, qr = to; // WA.
update(1, 1, N, add);
}
}
}
return 0;
}
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