您当前的位置: New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell(i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to celli using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
8 4 1 2 1 2 1 2 1
YES
8 5 1 2 1 2 1 1 1
NO
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
题意：给出n和m，给出1~n-1位置的跳跃值a[i]，位置t的初始值为1，每到达一个位置i，就可以往后跳跃到t + a[i]的位置上，依次循环，直到把所有的位置都跳过，注意，只能向后跳。问从开始能否恰好跳到位置m。
解析：直接暴力处理，看是否恰好走到即可。
AC代码：
非数组版的：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
#define INF 0x7fffffff
#define LL long long
#define LLL(a, b) a<<b
#define RRR(a, b) a>>b
#define MID(a, b) a+(b-a)/2
const int maxn = 30000 + 10;
int a[maxn];
int main(){
#ifdef sxk
freopen("in.txt", "r", stdin);
#endif // sxk
int n, m, x;
while(scanf("%d%d", &n, &m)!=EOF){
int t = 1, flag = 0;
for(int i=1; i<n; ){
if(t == m) flag = 1;
scanf("%d", &x);
t += x;
int y = x - 1;
while(y--) scanf("%d", &x);
i = t;
}
if(n == m) flag = 1;
printf("%s\n", flag ? "YES" : "NO");
}
return 0;
}
数组版的：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
#define INF 0x7fffffff
#define LL long long
#define LLL(a, b) a<<b
#define RRR(a, b) a>>b
#define MID(a, b) a+(b-a)/2
const int maxn = 30000 + 10;
int a[maxn];
int main(){
#ifdef sxk
freopen("in.txt", "r", stdin);
#endif // sxk
int n, m, x;
while(scanf("%d%d", &n, &m)!=EOF){
int t = 1, flag = 0;
for(int i=1; i<n; i++) scanf("%d", &a[i]);
for(int i=1; i<n; ){
t += a[i];
i = t;
if(t == m){
flag = 1;
break;
}
}
printf("%s\n", flag ? "YES" : "NO");
}
return 0;
}