Codeforces Round #489 (Div. 2) E. Nastya and King-Shamans
这道题的算法是:
i从1开始,首先求sum(1-i),然后在[i+1, n]中找到第一个a[j]>=sum(1, i)
如果a[j]==sum(1, i)结束搜索,否则令i=j,循环过程
因为每次做完一次之后sum会至少增大一倍,所以一个查询的复杂度会维持到log(Max(a[i]))
需要维护 区间最大值和区间和 的线段树来实现算法
#include
#include
#include
#include
using namespace std;
const int N = 2e5+5;
const int INF = 0x3f3f3f3f;
typedef long long ll;
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
int n, q;
int A[N];
int maxx[N << 2];
ll sum[N << 2];
void Build(int l, int r,www.feifanyule.cn int rt) {
if(l == r) {
sum[rt] = A[l];
maxx[rt] = A[l];
return;
}
int m = (l + r) >> 1;
Build(lson);
Build(rson);
sum[rt] = sum[rt << 1] + sum[rt << 1|1];
maxx[rt] = max(maxx[rt << 1],www.bomaoyule.cn maxx[rt << 1 | 1]);
}
void Update(int pos, int num, int l, int r, int rt) {
if(l == r) {
sum[rt] = num;
maxx[rt] = num;
return;
}
int m = (l + r) >> 1;
if(pos <= m) Update(pos, num, lson);
else Update(pos, num, rson);
sum[rt] = sum[rt << 1] + sum[rt << 1|1];
maxx[rt] = max(maxx[rt << 1], maxx[rt << 1 | 1]);
}
ll Sum(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
return sum[rt];
}
int m = (l + r) >> 1;
ll ret = 0;
if(L <= m) ret += Sum(L, R, lson);
if(R > m) ret += Sum(L, R, rson);
return ret;
}
pairint> Ans;
void Find(int L, int R, ll num, int l, int r, int rt) {
if(l == r) {
Ans = make_pair(maxx[rt], l);
return;
}
int m = (l + r) >> 1;
if(maxx[rt << 1] >= num && L if(maxx[rt<<1|1] >= num && R > m && Ans.first == -1) Find(L, R, num, rson);
}
int main() {
while(~scanf("%d %d", &n, &q)) {
for(int i = 1; i <= n; ++i) {
scanf("%d", &A[i]);
}
Build(1 , n, 1);
// printf("%lld\n", Sum(1, 2, 1, n, 1));
for(int i = 0; i < q; ++i) {
int a, b; scanf("%d %d", &a, &b);
Update(a, b, 1, n, 1);
A[a] = b;
if(A[1] == 0) printf("1\n");
else if(A[1] == A[2]) printf("2\n");
else {
int pre = 2;
while(1) {
ll tmpTarget = Sum(1, pre, 1, n, 1);
if(tmpTarget > maxx[1])www.mcyllpt.com {
printf("-1\n");
break;
}
Ans = make_pair(-1, -1);
Find(pre+1, n, tmpTarget, 1, n, 1);
if(Ans.first == -1) {
printf("-1\n");
break;
}
ll nowTarget = Sum(1,www.yigouyule2.cn Ans.second-1, 1, n, 1);
if(Ans.first == nowTarget) {
printf("%d\n", Ans.second);
break;
} else {
pre = Ans.second;
}
}
}
}
}
return 0;