codeforces 797E Array Queries

题目链接：http://codeforces.com/problemset/problem/797/E

题意：给定一个数组a[n]，q次询问，给定p、k，每次使p = p + a[p] + k，回答p超过n时的次数。n, q 小于 100000，p、k小于n，a[i] 小于n。

题解：暴力显然过不了极端情况，考虑到在k较大时，暴力跑时次数会很少，则对k进行分类，大于MAXK时直接暴力，小于时记忆化即可。

代码：

#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <cstdio>#include <cstdlib>#include <vector>#include <stack>#include <set>#include <queue>#include <functional>#include <map>#include <bitset>#define INF 0x7fffffff#define REP(i,j,k) for(int i = j;i <= k;i++)#define squr(x) (x) * (x)#define lowbit(x) (x&(-x))#define getint(x) scanf("%d", &(x))using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e5 + 10;const int MAXK = 200;int n, a[MAXN], q, p, k;int dp[MAXN][MAXK];int ans;int dfs(int x) {    if (x > n) {        return 0;    }    if (dp[x][k] != 0) {        return dp[x][k];    }    return dp[x][k] = 1 + dfs(x + a[x] + k);}int main(int argc, const char * argv[]) {    getint(n);    REP(i, 1, n) {        getint(a[i]);    }    getint(q);    REP(i, 1, q) {        ans = 0;        getint(p);        getint(k);        if (k >= MAXK) {            while (p <= n) {                p = p + k + a[p];                ans++;            }            printf("%d\n", ans);        } else {            printf("%d\n", dfs(p));        }    }    return 0;}