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6 3 1 4 2 10 3 10 2 16 4 16 5
YES YES NO YES YES NO
In the first test case, you can represent 3 as 3.
In the second test case, the only way to represent 4 is 1+3.
In the third test case, you cannot represent 10 as the sum of three distinct positive odd integers.
In the fourth test case, you can represent 10 as 3+7, for example.
In the fifth test case, you can represent 16 as 1+3+5+7.
In the sixth test case, you cannot represent 16 as the sum of five distinct positive odd integers.
题意：输入一个数n，让你判断是否存在k个奇数，使这k个奇数相加等于n。
思路：本题主要用了数学知识，再加上那么一些思考。
数学知识：1.奇+奇=偶；偶+偶=偶；奇+偶=奇；
2.前n个奇数的和为n²，即n个不同的奇数相加最小值为n²。
应用到本题是这样，k为奇数，n为奇数，那么k个奇数相加一定为奇数，符合题意
k为偶数，n为偶数，那么k个偶数相加一定为偶数，符合题意
然后剩下的情况中，由数学知识2知，n不能大于k*k，否则一定无解。
代码如下：
1 #include <iostream> 2 #include<cstdio> 3 using namespace std; 4 typedef long long ll; 5 int main() 6 { 7 ll t,n,k; 8 cin>>t; 9 while(t--) 10 { 11 scanf("%I64d%I64d",&n,&k); 12 if(n<k*k) 13 printf("NO\n"); 14 else 15 { 16 if((n&1)==(k&1))//若n和k同为奇或偶，则符合条件 17 printf("YES\n");//“与”运算判断奇偶 18 else 19 { 20 printf("NO\n"); 21 } 22 } 23 } 24 return 0; 25 }
CodeForces - 1327A Sum of Odd Integers（数学+思维）
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原文地址：https://www.cnblogs.com/theshorekind/p/12568364.html
