LOJ6280 數列分塊入門 4
標籤
- 分塊入門
前言
- 我的csdn和博客園是同步的,歡迎來訪danzh-博客園~
簡明題意
- 維護序列,支持兩種操作:
- 區間加
- 區間查詢
思路
- 多維護一個tag[]和一個sum[]就可以了~
注意事項
- 無
總結
- 無
AC代碼
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int maxn = 1e5 + 10;
int n, a[maxn];
int pos[maxn], len, tag[maxn], sum[maxn];
void change(int l, int r, int c)
{
for (int i = l; i <= min(len * pos[l], r); i++)
a[i] += c, sum[pos[i]] += c;
if (pos[l] != pos[r])
for (int i = r; i >= len * pos[r] - len + 1; i--)
a[i] += c, sum[pos[i]] += c;
for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
tag[i] += c;
}
int cal(int l, int r, int c)
{
long long ans = 0;
for (int i = l; i <= min(len * pos[l], r); i++)
ans += 1ll * a[i] + tag[pos[i]], ans %= (c + 1);
if (pos[l] != pos[r])
for (int i = r; i >= len * pos[r] - len + 1; i--)
ans += 1ll * a[i] + tag[pos[i]], ans %= (c + 1);
for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
ans += sum[i] + 1ll * tag[i] * len, ans %= (c + 1);
return ans;
}
void solve()
{
scanf("%d", &n);
len = sqrt(n);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]), pos[i] = (i - 1) / len + 1, sum[pos[i]] += a[i];
for (int i = 1; i <= n; i++)
{
int opt, l, r, c;
scanf("%d%d%d%d", &opt, &l, &r, &c);
if (opt == 0)
change(l, r, c);
else
printf("%d\n", cal(l, r, c));
}
}
int main()
{
freopen("Testin.txt", "r", stdin);
//freopen("Testout.txt", "w", stdout);
solve();
return 0;
}