在从小到大的排序数组中，
lower_bound( begin,end,num)：从数组的begin位置到end-1位置二分查找第一个大于或等于num的数字，找到返回该数字的地址，不存在则返回end。通过返回的地址减去起始地址begin,得到找到数字在数组中的下标。
upper_bound( begin,end,num)：从数组的begin位置到end-1位置二分查找第一个大于num的数字，找到返回该数字的地址，不存在则返回end。通过返回的地址减去起始地址begin,得到找到数字在数组中的下标。
unique函数返回第一个重复元素的位置，必须排序完才有用。
在从大到小的排序数组中，重载lower_bound()和upper_bound()。
lower_bound( begin,end,num,greater<type>() ):从数组的begin位置到end-1位置二分查找第一个小于或等于num的数字，找到返回该数字的地址，不存在则返回end。通过返回的地址减去起始地址begin,得到找到数字在数组中的下标。
upper_bound( begin,end,num,greater<type>() ):从数组的begin位置到end-1位置二分查找第一个小于num的数字，找到返回该数字的地址，不存在则返回end。通过返回的地址减去起始地址begin,得到找到数字在数组中的下标。
注意事项：如果不存在，则返回end - begin;
set中用greater表示从大到小排序，有限队列里面用graeter表示小的先出来。
#include<bits/stdc++.h> using namespace std; int main() { set<int, greater<int> > s; s.insert(1); s.insert(2); set<int, greater<int> > :: iterator it; for(it = s.begin(); it != s.end(); it++) { printf("%d ", *it); } priority_queue<int, vector<int>, greater<int> > q; q.push(1); q.push(2); printf("%d ", q.top()); return 0; }
multiset的使用
https://blog.csdn.net/sodacoco/article/details/84798621
string 中erase使用
https://www.cnblogs.com/yaoyueduzhen/p/4366267.html
struct rec{ int x,y; }; struct cmp{ bool operator()(const rec&a,const rec&b){ return a.x<b.x||a.x==b.x&&a.y<b.y; } }; multiset<rec,cmp>h;
struct Point{
int a,b;
bool operator <(const Point& rhs)const{
return a<rhs.a||(a==rhs.a && b <rhs.b);
}
};
multiset<Point> S;
multiset<Point>::iterator it;
binary_search(arr[],arr[]+size , indx) 二分查找函数，找到返回1，没找到返回0.
set中的lower_bound与upper_boound函数用法，equal_range函数是看这个数组内有多少相同的元素。
// g++ -o setLowerUpperBoundTest setLowerUpperBoundTest.cpp #include <set> #include <iostream> using namespace std ; typedef set<int> SET_INT; int main() { SET_INT s1; SET_INT::iterator i; cout << "s1.insert(5)" << endl; s1.insert(5); cout << "s1.insert(10)" << endl; s1.insert(10); cout << "s1.insert(15)" << endl; s1.insert(15); cout << "s1.insert(20)" << endl; s1.insert(20); cout << "s1.insert(25)" << endl; s1.insert(25); cout << "s1 -- starting at s1.lower_bound(12)" << endl; // prints: 15,20,25 for (i=s1.lower_bound(12);i!=s1.end();i++) cout << "s1 has " << *i << " in its set." << endl; cout << "s1 -- starting at s1.lower_bound(15)" << endl; // prints: 15,20,25 for (i=s1.lower_bound(15);i!=s1.end();i++) cout << "s1 has " << *i << " in its set." << endl; cout << "s1 -- starting at s1.upper_bound(12)" << endl; // prints: 15,20,25 for (i=s1.upper_bound(12);i!=s1.end();i++) cout << "s1 has " << *i << " in its set." << endl; cout << "s1 -- starting at s1.upper_bound(15)" << endl; // prints: 20,25 for (i=s1.upper_bound(15);i!=s1.end();i++) cout << "s1 has " << *i << " in its set." << endl; cout << "s1 -- s1.equal_range(12)" << endl; // does not print anything for (i=s1.equal_range(12).first;i!=s1.equal_range(12).second;i++) cout << "s1 has " << *i << " in its set." << endl; cout << "s1 -- s1.equal_range(15)" << endl; // prints: 15 for (i=s1.equal_range(15).first;i!=s1.equal_range(15).second;i++) cout << "s1 has " << *i << " in its set." << endl; } /* Output in windows s1.insert(5) s1.insert(10) s1.insert(15) s1.insert(20) s1.insert(25) s1 -- starting at s1.lower_bound(12) s1 has 15 in its set. s1 has 20 in its set. s1 has 25 in its set. s1 -- starting at s1.lower_bound(15) s1 has 15 in its set. s1 has 20 in its set. s1 has 25 in its set. s1 -- starting at s1.upper_bound(12) s1 has 15 in its set. s1 has 20 in its set. s1 has 25 in its set. s1 -- starting at s1.upper_bound(15) s1 has 20 in its set. s1 has 25 in its set. s1 -- s1.equal_range(12) s1 -- s1.equal_range(15) s1 has 15 in its set. */
#include <cstring> #include <cstdio> using namespace std; char a[7][10] = {"sunday", "monday", "tuesday", "wednesday", "thursday", "friday", "saturday"}; int main() { int y, m, d; while(scanf("%d %d %d", &y, &m, &d) != EOF) { if(m < 3) { y--; m += 12; } int c = y / 100; y = y % 100; int w = y + y/4 + c/4 - 2*c + 26 * (m+1)/10 + d - 1; printf("%s ", a[(w%7+7)%7]); } return 0; }
void init() { pw[0] = 1; for(int i = 1; i < 15; i++) { pw[i] = pw[i - 1] * 10; } }
while (next_permutation(str.begin(), str.end())) { k++; cout << str << endl; if (k == 5) { break; }
原理1： 把多于n+k个的物体放到n个抽屉里，则至少有一个抽屉里的东西不少于两件。
原理2 ：把多于mn(m乘以n)+1（n不为0）个的物体放到n个抽屉里，则至少有一个抽屉里有不少于（m+1）的物体。
原理3 ：把无穷多件物体放入n个抽屉，则至少有一个抽屉里 有无穷个物体。
第二抽屉原理
把（mn－1）个物体放入n个抽屉中，其中必有一个抽屉中至多有（m—1）个物体(例如，将3×5-1=14个物体放入5个抽屉中，则必定有一个抽屉中的物体数少于等于3-1=2)。 --------------------- 本文来自 数数1234 的CSDN 博客 ，全文地址请点击：https://blog.csdn.net/l2580258/article/details/51052631?utm_source=copy
bool palindromes(int start, int enda){ for(int i = start, j = enda; i < j; i++, j--) { if(ch[i] != ch[j]) { return false; } } return true; }
10.atan2与atan函数使用
printf("%lf %lf %lf %lf
", atan2(1, 1), atan2(-1, 1), atan2(-1, -1), atan2(1, -1));
printf("%lf %lf %lf %lf
", atan(1.0/1), atan(-1.0/1), atan(-1.0/-1), atan(1.0/-1));
主要是int_128的输入与输出。
#include <iostream> #define ll __int128 using namespace std; inline __int128 read() { __int128 x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } return x * f; } inline void print(__int128 x) { if (x < 0) { putchar('-'); x = -x; } if (x > 9) print(x / 10); putchar(x % 10 + '0'); } ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ll d = exgcd(b, a % b, x, y); ll z = x; x = y; y = z - y * (a / b); return d; } int main() { int n; scanf("%d",&n); ll val=read(); ll a, b, x, y, k; ll lcm = a, ans = b; bool flag = 1; --n; while (n--) { a=read(),b=read(); b = (b - ans % a + a) % a; ll d = exgcd(lcm, a, x, y); if (b % d) flag = 0; else k = x * (b / d) % a; ans += k * lcm; lcm = lcm / d * a; ans = (ans % lcm + lcm) % lcm; } if(!flag) puts("he was definitely lying"); else if(ans>val) puts("he was probably lying"); else print(ans); return 0; }