hdu 5084 前缀和预处理

http://acm.hdu.edu.cn/showproblem.php?pid=5084

给出矩阵M，求M*M矩阵的r行c列的数，每个查询跟前一个查询的结果有关。

观察该矩阵得知，令ans = M*M，则 ans[x][y] = (n-1-x行的每个值)*(n-1+y列的每个值)，即：

ans[x][y] = t[y] * t[2*n - 2 - x] +....+ t[y + n - 1]*t[n - 1 - x]

每一对的和为定值2*n-2-x-y，然后就是求每对i+j的前缀和(所有i+j值相同按顺序加起来)

#pragma comment(linker, "/STACK:36777216")
#pragma GCC optimize ("O2")
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
const int modo = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 1005,maxm = 1e4 + 5;
int t[maxn<<1],ans[maxn<<1][maxn<<1],n,m;

int main(){
    int x,y,_n;
    while(~RD(n)){
        _n = n;
        n = -2 + (n<<1);
        clr0(ans);
        for(int i = 0;i <= n;++i)
            RD(t[i]);
        for(int i = 0;i <= n;++i)
            for(int j = 0;j <= n;++j)
            ans[i][j] = t[i]*t[j];

        for(int i = 1;i <= n;++i)
            for(int j = 0;j < n;++j){
                ans[i][j] += ans[i-1][j+1];
            }
        RD(m);
        LL sum = 0;int res = 0;
        while(m--){
            RD2(x,y);
            x = (x+res)%_n,y = (y+res)%_n;
            int sx = _n - 1 + y,sy = _n - 1 - x;
            res = ans[sx][sy];
            if(sx - _n >= 0 && sy + _n <= n)
                res -= ans[sx-_n][sy+_n];
            sum += res;
//            cout<<x<<','<<y<<endl;
//            cout<<res<<endl;
        }
        printf("%I64d
",sum);
    }
    return 0;
}