给出仅含基本操作（
op）及循环语句（loop…end、continue、break）之程序，求出所含基本操作次数之多项式。基本操作与循环次数可能取正整数或 n。
由于无条件语句，事实上 continue 相当于 end loop 0，break 亦然，惟循环次数须设为 1，由此即可得解。
可惜，简单模拟我仍交了四回。其原因在 continue 或 break 时保留原循环以匹配其对应 end，同时循环次数置 0 从而忽略此后语句，但若此后再有 break 时如前所述又会置 1，从而出错。修改 (1) 处，在已置 0 时不再置 1，从而通过。
#include <cstring>
#include <cstdio>
const int oxis = 51;
struct status {
int loop;
int coe[oxis];
} zhan[oxis];
int get_num()
{
char now[oxis];
int result = -1;
scanf("%s", now);
if (strcmp(now, "n"))
sscanf(now, "%d", &result);
return result == 0 ? -1 : result; // 此处为修正数据错误
}
int main()
{
scanf("%*s");
int top = 0;
zhan[0].loop = 1;
while (true) {
char now[oxis];
scanf("%s", now);
if (strcmp(now, "loop") == 0) {
++top;
zhan[top].loop = get_num();
memset(zhan[top].coe, 0, sizeof(zhan[top].coe));
} else if (strcmp(now, "op") == 0) {
int cishu = get_num();
if (cishu == -1)
++zhan[top].coe[1];
else
zhan[top].coe[0] += cishu;
} else {
bool is_end = (strcmp(now, "end") == 0);
bool is_break = (strcmp(now, "break") == 0);
if (top == 0) {
if (is_end)
break;
else
continue;
}
if (is_break && zhan[top].loop != 0) // (1)
zhan[top].loop = 1;
int xin = top - 1, duo = zhan[top].loop;
if (duo == -1)
for (int i = 0; i < oxis - 1; ++i)
zhan[xin].coe[1 + i] += zhan[top].coe[i];
else
for (int i = 0; i < oxis; ++i)
zhan[xin].coe[i] += zhan[top].coe[i] * duo;
if (is_end)
top = xin;
else
zhan[top].loop = 0;
}
}
bool first = true;
for (int i = oxis - 1; i > 1; --i) {
int xs = zhan->coe[i];
if (xs == 0)
continue;
if (!first)
putchar('+');
if (xs != 1)
printf("%d", xs);
printf("n^%d", i);
first = false;
}
int xs = zhan->coe[1];
if (xs != 0) {
if (!first)
putchar('+');
if (xs != 1)
printf("%d", xs);
putchar('n');
first = false;
}
xs = zhan->coe[0];
if (xs != 0) {
if (!first)
putchar('+');
printf("%d", xs);
} else if (first)
putchar('0');
puts("");
return 0;
}