这道题很简单，直接算就可以了，直白模拟数学题。 关键是输入输出相当不好搞。
// 3495262 2013-12-23 23:18:43 Accepted 3504 C++ 0 168 呆滞的慢板
include
include
include
int main() {
int i, j, n;
double r1[16], r2[16], i1[16], i2[16], p, ans;
char s1[257], s2[257], *ss1, *ss2;
while(!feof(stdin)) {
gets(ss1 = s1);
gets(ss2 = s2);
scanf("%lf ", &p);
for(n = 0; *ss1; ++n) {
sscanf(ss1, " (%lf,%lf)", r1+n, i1+n);
sscanf(ss2, " (%lf,%lf)", r2+n, i2+n);
ss1 = strchr(ss1, ')')+1;
ss2 = strchr(ss2, ')')+1;
}
ans = 0;
for(i = 0; i < n; ++i) {
ans += pow((r1[i]-r2[i])*(r1[i]-r2[i])+(i1[i]-i2[i])*(i1[i]-i2[i]), 0.5*p);
}
printf("%.10lf\n", pow(ans, 1.0/p));
}
}
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原文链接：http://www.huangwenchao.com.cn/2013/12/zoj3504.html【zoj 3504 – P-norm】