HDU 6336 Matrix from Arrays (杭电多校4E)
遇事不决先打表。
然后会发现(个屁)大的矩形是由一个2L*2L的矩形重复出现组成的然后我们就可以这个矩形分成四个点到(0, 0)点的矩形,这样问题就变成了求四个到顶点(0, 0)的矩形的面积,然后就先去求这里面完整的块数,然后去找边缘的有一边是完整的块,然后找最右下角的没有完整的块的面积,然后加起来就可了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<map> #include<set> #include<ctime> #include<cmath> #include<stack> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define lowbit(x) (x & (-x)) typedef unsigned long long int ull; typedef long long int ll; const double pi = 4.0*atan(1.0); const int inf = 0x3f3f3f3f; const int maxn = 1e5+10; const int maxm = 10000007; const int mod = 1e9+7; using namespace std; int n, m, tol, T; int L; int A[15]; int M[100][100]; ll sx[30]; ll sy[30]; ll sum; void init() { sum = 0; memset(A, 0, sizeof A); memset(M, 0, sizeof M); memset(sx, 0, sizeof sx); memset(sy, 0, sizeof sy); } ll solve(int x, int y) { ll ans = 0; int cx = x/L; int cy = y/L; ll t = 1ll * cx * cy; ans += 1ll * t * sum; int x1 = x % L; int y1 = y % L; ll s = 0; for(int i=1; i<=x1; i++) s += sx[i]; ans += s * cy; s = 0; for(int i=1; i<=y1; i++) s += sy[i]; ans += s * cx; for(int i=1; i<=x1; i++) { for(int j=1; j<=y1; j++) { ans += M[i][j]; } } return ans; } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while(T--) { init(); scanf("%d", &L); for(int i=0; i<L; i++) scanf("%d", &A[i]); int cursor = 0; for(int i=1; i<=4*L; i++) { for(int j=1; j<=i; j++) { M[j][i - j + 1] = A[cursor]; cursor = (cursor + 1) % L; } } L <<= 1; for(int i=1; i<=L; i++) { for(int j=1; j<=L; j++) { sy[j] += M[i][j]; sx[i] += M[i][j]; } sum += sx[i]; } /* for(int i=1; i<=L; i++) for(int j=1; j<=L; j++) printf("%d%c",M[i][j], j==L ? '\n' : ' \t'); for(int i=1; i<=L; i++) printf("sx%d = %d\n", i, sx[i]); for(int i=1; i<=L; i++) printf("sy%d = %d\n", i, sy[i]); */ int q; scanf("%d", &q); while(q--) { int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); x1++, y1++, x2++, y2++; ll ans = solve(x2, y2); ans -= solve(x1-1, y2); ans -= solve(x2, y1-1); ans += solve(x1-1, y1-1); printf("%lld\n", ans); } } return 0; }