[expelliarmus] the disarming charm [rictusempra] send a jet of silver light to hit the enemy [tarantallegra] control the movement of one‘s legs [serpensortia] shoot a snake out of the end of one‘s wand [lumos] light the wand [obliviate] the memory charm [expecto patronum] send a Patronus to the dementors [accio] the summoning charm @END@ 4 [lumos] the summoning charm [arha] take me to the skySample Output
light the wand accio what? what?
题意 ： 给你一个咒语对应着一个功能，要求有 q 此询问，对于每次询问给出相应的咒语后告诉你相应的功能
思路分析：这题用 map 可能会超内存，因此我们这里用 hash 去做，对于 hash 后的值相同的情况下我们去连一条链，然后匹配的时候去在这条链上去匹配即可。
代码示例：
using namespace std;
#define ll unsigned long long
const ll maxn = 1e6+5;
char s[200];
struct node
{
char que[25];
char ans[85];
int next;
}a[maxn], b[maxn];
ll p = 100007;
char que_[25], ans_[85];
int ha[maxn], hb[maxn];
ll cura=0, curb=0;
ll gethash(char *str){
ll res = 0;
for(ll i = 0; *(str+i); i++){
res = res*p+(str[i]-‘a‘);
}
return res%p;
}
void insert(){
ll hash_a = gethash(que_);
strcpy(a[cura].que, que_);
strcpy(a[cura].ans, ans_);
a[cura].next = ha[hash_a];
ha[hash_a] = cura;
cura++;
ll hash_b = gethash(ans_);
strcpy(b[curb].que, que_);
strcpy(b[curb].ans, ans_);
b[curb].next = hb[hash_b];
hb[hash_b] = curb;
curb++;
//printf("++++ %llu %llu \n", hash_a, hash_b);
}
bool searcha(char *str){
ll num = gethash(str);
int x= ha[num];
//printf("1111111111 %llu %d\n", num, x);
while(x != -1){
//printf("_______ %s \n", a[x].que);
if (strcmp(a[x].que, str) == 0){
printf("%s\n", a[x].ans);
return true;
}
x = a[x].next;
}
return false;
}
bool searchb(char *str){
ll num = gethash(str);
int x= hb[num];
//printf("2222222222 %llu %llu\n", num, x);
while(x != -1){
if (strcmp(b[x].ans, str) == 0){
printf("%s\n", b[x].que);
return true;
}
x = b[x].next;
}
return false;
}
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
memset(ha, -1, sizeof(ha));
memset(hb, -1, sizeof(hb));
while(1){
gets(s+1);
ll len = strlen(s+1);
ll pos = 1;
if (s[1] == ‘@‘) break;
ll k = 0;
for(ll i = 2; i <= len; i++){
if (s[i] == ‘]‘) {pos = i; break;}
que_[k++] = s[i];
}
que_[k] = ‘\0‘;
k = 0;
for(ll i = pos+2; i <= len; i++){
ans_[k++] = s[i];
}
ans_[k] = ‘\0‘;
insert();
}
ll q;
cin >>q;
getchar();
while(q--){
gets(s);
ll len = strlen(s);
if (s[0] == ‘[‘){
s[len-1] = ‘\0‘;
s[0] = ‘\0‘;
if (!searcha(s+1)) printf("what?\n");
}
else {
if (!searchb(s)) printf("what?\n");
}
}
return 0;
}
原文：https://www.cnblogs.com/ccut-ry/p/9380592.html