#include <iostream>
using namespace std;
int main() {int n;scanf("%d", &n);for (int i = 0; i < n; i++) {long long int a, b, c;scanf("%lld%lld%lld", &a, &b, &c);printf("Case #%d: %s\n", i + 1, a + b > c ? "true" : "false");}return 0;
}

#include <iostream>
using namespace std;
// 這道題目，真抽象，怪不得通過率這么低，壓根是因為邊界測試緣故，粗心緣故
int main() {int A1=0,A2=0,A3=0,A4=0,A5=INT32_MIN;int i2=1,i4=0;int n;cin>>n;int num;while(n--){cin>>num;if(num%5==0&&num%2==0){A1 += num;}else if (num%5==1){A2 += num*i2;i2*=(-1);}else if (num%5==2) {A3++;}else if (num%5==3){A4+=num;i4++;}else if (num%5==4) {A5 = max(num,A5);}}if(A1!=0) cout<<A1<<" ";else cout<<"N ";if(A2!=0) cout<<A2<<" ";else cout<<"N ";if(A3!=0) cout<<A3<<" ";else cout<<"N ";if(i4!=0) printf("%.1f ",(double)A4/i4);else cout<<"N ";if(A5!=INT32_MIN) cout<<A5<<" ";else cout<<"N ";
}
// 64 位輸出請用 printf(\"%lld\")

#include "bits/stdc++.h"
using namespace std;
int sieve[110005];
void get_sieve(vector<bool>& bool_sieve){ // 歐拉篩，大范圍篩數int num = 0;bool_sieve[2] = true; // 表示二是素數for(int i=2; i<110005; ++i){if(bool_sieve[i]) sieve[num++]=i;for (int j = 0; j < num; ++j) {if(i*sieve[j]>110005) break;bool_sieve[i*sieve[j]] = false;if(i% sieve[j]==0) break;}}}
int main(){int n,m;cin>>n>>m;vector<bool> bool_sieve(110005, true); // 表示都是質數bool_sieve[0] = bool_sieve[1] = false;get_sieve(bool_sieve);int flag = 0;for(int i = n-1; i < m; ++i){flag++;cout<<sieve[i];if(flag%10==0) cout<<endl;else cout<<" ";}return 0;
}

#include <iostream>
#include <cctype>
using namespace std;
// 這個是劉姐的代碼，太簡介啦！！！
int main() {string a, b, c, d;cin >> a >> b >> c >> d;char t[2];int pos, i = 0, j = 0;while(i < a.length() && i < b.length()) {if (a[i] == b[i] && (a[i] >= 'A' && a[i] <= 'G')) {t[0] = a[i];break;}i++;}i = i + 1;while (i < a.length() && i < b.length()) {if (a[i] == b[i] && ((a[i] >= 'A' && a[i] <= 'N') || isdigit(a[i]))) {t[1] = a[i];break;}i++;}while (j < c.length() && j < d.length()) {if (c[j] == d[j] && isalpha(c[j])) {pos = j;break;}j++;}string week[7] = {"MON ", "TUE ", "WED ", "THU ", "FRI ", "SAT ", "SUN "};int m = isdigit(t[1]) ? t[1] - '0' : t[1] - 'A' + 10;cout << week[t[0]-'A'];printf("%02d:%02d", m, pos);return 0;
}

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
// 這可是，我第一次，在算法中運用結構體，太爽了，當與cmp結合時
struct student {int sum;int moral;int talent;int id;student(int x, int y, int z): moral(x), talent(y), id(z) {sum = x + y;}
};
bool cmp(const student& stu1, const student& stu2) { // 匹配模式if (stu1.sum != stu2.sum) return stu1.sum >stu2.sum; // 第一個大于第二個if (stu1.moral != stu2.moral) return stu1.moral > stu2.moral; // 德高者居之return stu1.id < stu2.id; // 按降序排列
}
int main() {int N, L, H; // N(考生總數) L(最低錄取分數線) H(優先錄取)cin >> N >> L >> H;vector<student> vec1, vec2, vec3, vec4;while (N--) {int id, moral, talent;cin >> id >> moral >> talent;student s(moral, talent, id);if (s.moral >= H && s.talent >= H) vec1.push_back(s);else if (s.moral >= L && s.talent >= L) { // 都及格if (s.moral >= s.talent && s.moral >= H) vec2.push_back(s);else if (s.moral >= s.talent) vec3.push_back(s);else vec4.push_back(s);}}sort(vec1.begin(), vec1.end(), cmp);sort(vec2.begin(), vec2.end(), cmp);sort(vec3.begin(), vec3.end(), cmp);sort(vec4.begin(), vec4.end(), cmp);cout << vec1.size() + vec2.size() + vec3.size() + vec4.size() << endl;for (student s : vec1) cout << s.id << " " << s.moral << " " << s.talent <<endl;for (student s : vec2) cout << s.id << " " << s.moral << " " << s.talent <<endl;for (student s : vec3) cout << s.id << " " << s.moral << " " << s.talent <<endl;for (student s : vec4) cout << s.id << " " << s.moral << " " << s.talent <<endl;return 0;
}

#include <bits/stdc++.h>
#define ll long long
using namespace std;
// 這道題，倒是挺簡單的
int get_num(char d, int num){int D = d-'0';int all_num = 0;while (num--) {all_num = all_num*10+D;}return all_num;
}
int main(){string a,b;char Da,Db;cin>>a>>Da>>b>>Db; // 輸入int numa=0,numb=0;for(int i=0; i<a.size(); ++i) // 判斷if(a[i]==Da) numa++;for(int i=0; i<b.size(); ++i)if(b[i]==Db) numb++;cout<<get_num(Da, numa)+get_num(Db, numb)<<endl;}

// ！本題值得做！，主要模擬除法內容
// 加深對模擬的了解
#include <bits/stdc++.h>
using namespace std;
int main(){string a;int b;cin>>a>>b;int len,t=0,temp=0; // t(存放除數)，temp(存放余數)len = a.size();t = (a[0]-'0')/b; // 模擬除法if(t!=0) cout<<t;temp = (a[0]-'0')%b;int flag = false;for(int i=1; i<a.size(); ++i){t = (temp*10+a[i]-'0')/b; // 求除數if(t!=0) flag = true;if(flag) cout<<t;temp = (temp*10 +a[i]-'0')%b; // 求剩下的余數}cout<<" "<<temp;return 0;
}

// 本題，一般般，沒啥含金量
// 其實挺有含金量的，知識不知道如何將其表示出來
// 這是我第一次，通過pair<int,int>的方式展現出來
#include "iostream"
#include "vector"
#include "algorithm"
using namespace std;
bool cmp(const pair<char,int>& p1,const pair<char,int>& p2){if(p1.second!=p2.second) return p1.second>p2.second;return p1.first<p2.first; // 升序
}
int main(){ // 這是個啥玩意，天啊，簡直嘍// C代表“錘子”、J代表“剪刀”、B代表“布”，分別除下vector<pair<char,int>> vec1{{'C',0},{'J',0},{'B',0}};vector<pair<char,int>> vec2{{'C',0},{'J',0},{'B',0}};int n;cin>>n;int a=0,b=0,c=0;char k1,k2;// 只需要，3個字母就能將勝負表示while(n--){cin>>k1>>k2;if(k1==k2){ // 平局b++;}else if((k1=='C'&&k2=='J')||(k1=='J'&&k2=='B')||(k1=='B'&&k2=='C')){ // 甲勝a++;if(k1=='C') vec1[0].second++;if(k1=='J') vec1[1].second++;if(k1=='B') vec1[2].second++;// 把另一種方法補充上去}else if((k2=='C'&&k1=='J')||(k2=='J'&&k1=='B')||(k2=='B'&&k1=='C')){ // 乙勝c++;if(k2=='C') vec2[0].second++;if(k2=='J') vec2[1].second++;if(k2=='B') vec2[2].second++;}}sort(vec1.begin(),vec1.end(), cmp);sort(vec2.begin(),vec2.end(), cmp);cout<<a<<" "<<b<<" "<<c<<endl;cout<<c<<" "<<b<<" "<<a<<endl;cout<<vec1[0].first<<" "<<vec2[0].first<<endl;return 0;
}
// 當然以上是最麻煩的一種方法之一
// ..... 接下來看這個
/*using namespace std;
int main() {int n;cin >> n;int jiawin = 0, yiwin = 0;int jia[3] = {0}, yi[3] = {0};for (int i = 0; i < n; i++) {char s, t;cin >> s >> t;if (s == 'B' && t == 'C') {jiawin++;jia[0]++;} else if (s == 'B' && t == 'J') {yiwin++;yi[2]++;} else if (s == 'C' && t == 'B') {yiwin++;yi[0]++;} else if (s == 'C' && t == 'J') {jiawin++;jia[1]++;} else if (s == 'J' && t == 'B') {jiawin++;jia[2]++;} else if (s == 'J' && t == 'C') {yiwin++;yi[1]++;}}cout << jiawin << " " << n - jiawin - yiwin << " " << yiwin << endl << yiwin << " " << n - jiawin - yiwin << " " << jiawin << endl;int maxjia = jia[0] >= jia[1] ? 0 : 1;maxjia = jia[maxjia] >= jia[2] ? maxjia : 2;int maxyi = yi[0] >= yi[1] ? 0 : 1;maxyi = yi[maxyi] >= yi[2] ? maxyi : 2;char str[4] = {"BCJ"};cout << str[maxjia] << " " << str[maxyi];return 0;
}*/

#include <iostream>
#include <algorithm>
using namespace std;
// 夯實了，我對string-字符串的運用
bool cmp(char a, char b){ // 默認是從小到大，現在是從大到小return a>b;
}
int main(){ // string str,str1,str2,str3;cin>>str;int num1,num2,num; // while (1) {sort(str.begin(),str.end(),cmp);str1 = str;num1 = stoi(str1);sort(str.begin(),str.end());str2 = str;num2 = stoi(str2);  num = num1-num2;str = to_string(num);str.insert(0,4-str.size(),'0');if(num==0){cout<<str1<<" - "<<str2<<" = "<<"0000"<<endl;break;} else {cout<<str1<<" - "<<str2<<" = "<<str<<endl;if(num==6174) break;}}return 0;
}

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
// 對struct不在排擠，而是使用
// 認為細節扣到他腦子有坑，哭死
struct mooncake{int num;int money;double price;mooncake():num(0),money(0),price(0){} // 構造器
};bool cmp(mooncake m1, mooncake m2){ // 從小到大排序return m1.price>m2.price;
}
int main(){int n,d; // n種類，d市場cin>>n>>d;vector<mooncake> vec(n);for(int i=0; i<n; ++i)  cin>>vec[i].num;for(int i=0; i<n; ++i){ // 計算cin>>vec[i].money;vec[i].price = (double)vec[i].money/vec[i].num;}sort(vec.begin(),vec.end(),cmp); // 挺好int remain = d;double money = 0;for(int i=0; i<vec.size();  ++i){if(remain>=vec[i].num){money += vec[i].money;remain -= vec[i].num;}else{money += remain*vec[i].price;remain = 0;}if(remain<=0) break;}printf("%.2lf",money);return 0;
}

#include "iostream"
using namespace std;int main(){// 增刪改查string str = "123";cout<<str<<endl;// 增 - 插入str.insert(0,1,'0');str.insert(0,"aaa");cout<<str<<endl;// 刪 - 刪除str.erase(0,1); // 用的好，甚至可以刪除某個特定字符cout<<str<<endl;// 改 - 交換，反轉string str2 = "00";str2.swap(str); - 反轉reverse(s.begin(), s.end());cout<<str<<endl;// 查 - 截取片段，查字符string str3 = str2.substr(0,3); // 截取cout<<str3<<endl;cout<<str3.find('0')<<endl;if(str3.find('1')==string::npos) cout<<"你好"<<endl;return 0;
}