Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem D: Snakes and Ladders

Snakes and Ladders is a board game played on a 10 by 10 grid. The squares of the grid are numbered 1 to 100. Each player has a token which is placed on the square numbered 1 at the beginning of the game. Players take turns rolling a die which provides a random number between 1 and 6. After rolling, the player advances his or her token the number of squares shown on the die. If this would put the token past the square numbered 100, the token is advanced to 100. After advancing, if the token is on a square containing the bottom of a ladder, the token must be moved to the square containing the top of the ladder. Similarly, if the token is on a square containing the mouth of a snake, the token must be moved to the square containing the tail of the snake. No square contains more than one endpoint of any snake or ladder. The square numbered 100 does not contain the mouth of a snake or the bottom of a ladder. A player wins when his or her token reached the square numbered 100. At that point, the game ends.

Given the configuration of the snakes and ladders on a game board and a sequence of die rolls, you are to determine the positions of all the tokens on the game board. The sequence of die rolls need not be complete (i.e. it need not lead to any player winning). The sequence of die rolls may also continue after the game is over; in this case, the die rolls after the end of the game should be ignored.

Input Specification

The first line is the number of test cases to follow. The test cases follow, one after another; the format of each test case is the following:

The first line contains three positive integers: the number a of players, the number b of snakes or ladders, and the number c of die rolls. There will be no more than 1000000 players and no more than 1000000 die rolls. Each of the next b lines contains two integers specifying a snake or ladder. The first integer indicates the square containing the mouth of the snake or the bottom of the ladder. The second integer indicates the square containing the tail of the snake or the top of the ladder. The following c lines each contain one integer giving number rolled on the die.

Sample Input

1
3 1 3
4 20
3
4
5

Output Specification

For each player, output a single line containing a string of the form Position of player N is P., where N is replaced with the number of the player and P is replaced with the final position of the player.

Output for Sample Input

Position of player 1 is 20.
Position of player 2 is 5.
Position of player 3 is 6.

---

Ondřej Lhoták

題目描述：
蛇梯遊戲, 為了這題還特地去玩了一下, 否則還真的有點難理解, 遇到梯子可以跳格, 遇到蛇則會退格, 有點像是賞罰一樣, 然後擲骰子前進, 如果超過 100 算 100, 而當其中一名宣告勝利時, 遊戲變會結束。
已經知道骰子依序會骰出什麼, 模擬過程, 輸出最後停留的位置。

題目解法：

有可能會有多餘的骰子剩餘。

#include <stdio.h>
int pos[1000005];
inline int readchar() {
    const int N = 1048576;
    static char buf[N];
    static char *p = buf, *end = buf;
    if(p == end) {
        if((end = buf + fread(buf, 1, N, stdin)) == buf) return EOF;
        p = buf;
    }
    return *p++;
}
inline int ReadInt(int *x) {
    static char c, neg;
    while((c = readchar()) < '-')    {if(c == EOF) return 0;}
    neg = (c == '-') ? -1 : 1;
    *x = (neg == 1) ? c-'0' : 0;
    while((c = readchar()) >= '0')
        *x = (*x << 3) + (*x << 1) + c-'0';
    *x *= neg;
    return 1;
}
int main() {
    /*freopen("in.txt", "r+t", stdin);
    freopen("out.txt", "w+t", stdout);*/
    int testcase;
    int n, m, q, i;
    ReadInt(&testcase);
    //scanf("%d", &testcase);
    while(testcase--) {// n players
        //scanf("%d %d %d", &n, &q, &m);
        ReadInt(&n);
        ReadInt(&q);
        ReadInt(&m);
        int x, y, link[105] = {};
        for(i = 1; i <= n; i++)
            pos[i] = 1;
        while(q--) {
            //scanf("%d %d", &x, &y);
            ReadInt(&x);
            ReadInt(&y);
            link[x] = y;
        }
        int turn = 1, roll, winner = 0;
        while(m--) {
            ReadInt(&roll);
            //scanf("%d", &roll);
            pos[turn] += roll;
            if(pos[turn] >= 100)
                pos[turn] = 100;
            if(link[pos[turn]])
                pos[turn] = link[pos[turn]];
            if(pos[turn] >= 100)
                pos[turn] = 100, winner = 1;
            turn++;
            if(turn > n)    turn = 1;
            if(winner) {
                while(m--)
                    ReadInt(&roll);
                    //scanf("%d", &roll);
                break;
            }
        }
        for(i = 1; i <= n; i++)
            printf("Position of player %d is %d.\n", i, pos[i]);
    }
    return 0;
}