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 E: Logo 2

Logo is a programming language built around a turtle. Commands in the language cause the turtle to move. The turtle has a pen attached to it. As the turtle moves, it draw lines on the page. The turtle can be programmed to draw interesting pictures.

We are interested in making the turtle draw a picture, then return to the point that it started from. For example, we could give the turtle the following program:

fd 100 lt 120 fd 100 lt 120 fd 100

The command fd causes the turtle to move forward by the specified number of units. The command lt causes the turtle to turn left by the specified number of degrees. Thus the above commands cause the turtle to draw an equilateral triangle with sides 100 units long. Notice that after executing the commands, the turtle ends up in the same place as it started. The turtle understands two additional commands. The command bk causes the turtle to move backward by the specified number of units. The command rt causes the turtle to turn right by the specified number of degrees. The distances and angles in all commands are always non-negative integers.

Unfortunately, we have been messy in writing the program down, and cannot read our own writing. One of the numbers in the program is missing. Assuming the turtle ends up at the place that it started at the end of its journey, can you find the missing number?

Input Specification

The first line of input contains one integer specifying the number of test cases to follow. Each test case starts with a line containing one integer, the number of commands to follow. The commands follow, one on each line. Each test case will contain no more than 1000 commands. The argument of each command is either an integer or a question mark (?). There will be exactly one question mark in each test case.

Sample Input

1
5
fd 100
lt 120
fd ?
lt 120
fd 100

Output Specification

For each test case, output line containing a single integer n such that when the question mark in the program is replaced by n, the turtle ends up at the same point that it started from once the program completes. If the question mark is the argument of an lt or rt command, the angle in the output must be between 0 and 359 degrees, inclusive. The correct answer will always be an integer, and we guarantee that for every test case, there will be only one correct answer.

Output for Sample Input

100

---

Ondřej Lhoták

如果缺角度就直接窮舉，如果缺距離就忽略，最後算離圓心的距離。
假設從 (0, 0) 出發。

#include <stdio.h>
#include <math.h>
char cc[1005][10];
int cv[1005], n;
const double pi = acos(-1);
void getDegree(int pos) {
    int i, j, ret = 0;
    double dif = 0xffff;
    for(i = 0; i < 360; i++) {
        double theta = 0, x = 0, y = 0;
        cv[pos] = i;
        for(j = 0; j < n; j++) {
            if(cc[j][0] == 'f') {//forward
                x += cv[j]*cos(theta*pi/180);
                y += cv[j]*sin(theta*pi/180);
            } else if(cc[j][0] == 'b') {
                x -= cv[j]*cos(theta*pi/180);
                y -= cv[j]*sin(theta*pi/180);
            } else if(cc[j][0] == 'l') {
                theta += cv[j];
            } else {
                theta -= cv[j];
            }
        }
        if(fabs(x)+fabs(y) < dif) {
            dif = fabs(x)+fabs(y);
            ret = i;
        }
    }
    printf("%d\n", ret);
}
void getDist(int pos) {
    int i, j;
    double px = 0, py = 0, qx = 0, qy = 0;
    double theta = 0;
    for(i = 0; i < pos; i++) {
        if(cc[i][0] == 'f') {//forward
            px += cv[i]*cos(theta*pi/180);
            py += cv[i]*sin(theta*pi/180);
        } else if(cc[i][0] == 'b') {
            px -= cv[i]*cos(theta*pi/180);
            py -= cv[i]*sin(theta*pi/180);
        } else if(cc[i][0] == 'l') {
            theta += cv[i];
        } else {
            theta -= cv[i];
        }
    }
    for(i = pos+1; i < n; i++) {
        if(cc[i][0] == 'f') {//forward
            qx += cv[i]*cos(theta*pi/180);
            qy += cv[i]*sin(theta*pi/180);
        } else if(cc[i][0] == 'b') {
            qx -= cv[i]*cos(theta*pi/180);
            qy -= cv[i]*sin(theta*pi/180);
        } else if(cc[i][0] == 'l') {
            theta += cv[i];
        } else {
            theta -= cv[i];
        }
    }
    printf("%.0lf\n", hypot(px+qx, py+qy));

}
int main() {
    int testcase, i;
    scanf("%d", &testcase);
    char foo[10];
    while(testcase--) {
        scanf("%d", &n);
        int pos = 0;
        for(i = 0; i < n; i++) {
            scanf("%s %s", cc[i], foo);
            if(foo[0] == '?')   cv[i] = -1, pos = i;
            else
                sscanf(foo, "%d", &cv[i]);
        }
        if(cc[pos][0] != 'f' && cc[pos][0] != 'b')
            getDegree(pos);
        else
            getDist(pos);
    }
    return 0;
}