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

It's normal to feel worried and tense the day before a programming contest. To relax, you went out for a drink with some friends in a nearby pub. To keep your mind sharp for the next day, you decided to play the following game. To start, your friends will give you a sequence of N integers X₁, X₂,..., XN. Then, there will be K rounds; at each round, your friends will issue a command, which can be:

• a change command, when your friends want to change one of the values in the sequence; or
• a product command, when your friends give you two values I, J and ask you if the product X_IxX_I+1x ... xX_J-1xX_J is positive, negative or zero.

Since you are at a pub, it was decided that the penalty for a wrong answer is to drink a pint of beer. You are worried this could affect you negatively at the next day's contest, and you don't want to check if Ballmer's peak theory is correct. Fortunately, your friends gave you the right to use your notebook. Since you trust more your coding skills than your math, you decided to write a program to help you in the game.

Input 

Each test case is described using several lines. The first line contains two integers N and K, indicating respectively the number of elements in the sequence and the number of rounds of the game ( 1N, K10⁵). The second line contains N integers X_i that represent the initial values of the sequence ( -100X_i100 for i = 1, 2,..., N). Each of the next K lines describes a command and starts with an uppercase letter that is either ` C' or ` P'. If the letter is ` C', the line describes a change command, and the letter is followed by two integers I and V indicating that X<sub>I</sub> must receive the value V ( 1IN and -100V100). If the letter is ` P', the line describes a product command, and the letter is followed by two integers I and J indicating that the product from X_I to X_J, inclusive must be calculated ( 1IJN). Within each test case there is at least one product command.

Output 

For each test case output a line with a string representing the result of all the product commands in the test case. The i-th character of the string represents the result of the i-th product command. If the result of the command is positive the character must be ` +' (plus); if the result is negative the character must be ` -' (minus); if the result is zero the character must be ` 0' (zero).

Sample Input 

4 6-2 6 0 -1C 1 10P 1 4C 3 7P 2 2C 4 -5P 1 45 91 5 -2 4 3P 1 2P 1 5C 4 -5P 1 5P 4 5C 3 0P 1 5C 4 -5C 4 -5

Sample Output 

0+-+-+-0

---

線段樹－

記錄區間負的個數、以及有沒有零即可。

#include <stdio.h>
int A[100005];
int ST[262144][2];
void build(int k, int l, int r) {
    if(l == r) {
        ST[k][1] = (A[l] == 0);
        ST[k][0] = A[l] < 0;
        return ;
    }
    if(l < r) {
        int m = (l+r)/2;
        build(k<<1, l, m);
        build(k<<1|1, m+1, r);
        ST[k][0] = ST[k<<1][0]+ST[k<<1|1][0];
        ST[k][1] = ST[k<<1][1]|ST[k<<1|1][1];
    }
}
void update(int k, int l, int r, int qx, int qv) {
    if(l == r && l == qx) {
        A[l] = qv;
        ST[k][1] = (A[l] == 0);
        ST[k][0] = A[l] < 0;
        return;
    }
    int m = (l+r)/2;
    if(qx <= m)
        update(k<<1, l, m, qx, qv);
    else
        update(k<<1|1, m+1, r, qx, qv);
    ST[k][0] = ST[k<<1][0]+ST[k<<1|1][0];
    ST[k][1] = ST[k<<1][1]|ST[k<<1|1][1];
}
int neg, has;
void query(int k, int l, int r, int ql, int qr) {
    if(l == ql && r == qr) {
        neg += ST[k][0];
        has |= ST[k][1];
        return;
    }
    int m = (l+r)/2;
    if(qr <= m) {
        query(k<<1, l, m, ql, qr);
    } else if(ql > m) {
        query(k<<1|1, m+1, r, ql, qr);
    } else {
        query(k<<1, l, m, ql, m);
        query(k<<1|1, m+1, r, m+1, qr);
    }
}
int main() {
    int n, k, i, x, y;
    char cmd[2];
    while(scanf("%d %d", &n, &k) == 2) {
        for(i = 1; i <= n; i++)
            scanf("%d", &A[i]);
        build(1, 1, n);
        while(k--) {
            scanf("%s %d %d", cmd, &x, &y);
            if(cmd[0] == 'C') {
                update(1, 1, n, x, y);
            } else {
                neg = 0, has = 0;
                query(1, 1, n, x, y);
                if(has)
                    printf("0");
                else if(neg&1)
                    printf("-");
                else
                    printf("+");
            }
        }
        puts("");
    }
    return 0;
}