C. Star sky 二维前缀和
The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).
Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.
You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.
A star lies in a rectangle if it lies on its border or lies strictly inside it.
The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.
The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si(1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.
The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.
For each view print the total brightness of the viewed stars.
2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5
3
0
3
3 4 5
1 1 2
2 3 0
3 3 1
0 1 1 100 100
1 2 2 4 4
2 2 1 4 7
1 50 50 51 51
3
3
5
0
Let's consider the first example.
At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.
At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.
At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.
#include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cstring> #include<string> using namespace std; typedef long long LL; #define MAXN 109 #define N 100 /* 坐标的最大值比较小 开前缀数组求和 */ LL a[MAXN][MAXN][12]; LL n, q, c; int main() { cin >> n >> q >> c; int t1, t2, t3, t4; for (LL i = 0; i < n; i++) { cin >> t1 >> t2 >> t3; a[t1][t2][t3]++; } for(int i =1;i<=101;i++) for(int j=1;j<=101;j++) for (int k = 0; k < c + 1; k++) { a[i][j][k] += a[i - 1][j][k] + a[i][j - 1][k] - a[i - 1][j - 1][k]; } int T; int x1, y1, x2, y2; while (q--) { cin >> T >> x1 >> y1 >> x2 >> y2; LL cnt = 0, sum = 0; for (int i = 0; i < c + 1; i++) { cnt = a[x2][y2][i] - a[x2][y1 - 1][i] - a[x1 - 1][y2][i] + a[x1 - 1][y1 - 1][i]; sum += cnt*((i + T) % (c + 1)); } cout << sum << endl; } }
错误的二分代码:
#include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cstring> #include<string> using namespace std; typedef long long LL; #define MAXN 100005 #define N 100 /* 二分查找区域内符合范围的点,取模运算求和 ... */ struct star { LL x, y, s; bool operator<(const star& rhs) { if (x == rhs.x) return y < rhs.y; else return x < rhs.x; } }a[MAXN]; LL n, q, c; LL t, X1, Y1, X2, Y2; LL sum = 0; void solve(LL l, LL r) { if (l > r) return; if (l == r) { if (a[l].x >= X1&&a[l].x >= Y1&&a[l].x <= X2&&a[l].y <= Y2) sum += (a[l].s + t) % (c + 1); return; } LL mid = (l + r) / 2; //cout << mid << endl; if (a[mid].x > X2) solve(l, mid - 1); else if (a[mid].x < X1) solve(mid + 1, r); else { if (a[mid].x >= X1&&a[mid].x >= Y1&&a[mid].x <= X2&&a[mid].y <= Y2) sum += (a[mid].s + t) % (c + 1); solve(l, mid - 1); solve(mid + 1, r); } } int main() { scanf("%lld%lld%lld", &n, &q, &c); for (LL i = 0; i < n; i++) scanf("%lld%lld%lld", &a[i].x, &a[i].y, &a[i].s); sort(a, a + n); while (q--) { sum = 0; scanf("%lld%lld%lld%lld%lld", &t, &X1, &Y1, &X2, &Y2); solve(0, n-1); printf("%lld\n", sum); } return 0; }