codeforces 468A 24 Game 构造
2014-09-22 00:36
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题意:给定一个数n和“+”、“-”、“*”三种运算符号。问用[1,n]中所有的数能否构成24
思路:n<=3无解; n=4时,1*2*3*4=24; n=5时,4*2=8, 5+1-3=3 , 3*8=24; n>=6时,用1、2、3构成0,4、6构成24,其余全部和0乘即
可。详见代码:
思路:n<=3无解; n=4时,1*2*3*4=24; n=5时,4*2=8, 5+1-3=3 , 3*8=24; n>=6时,用1、2、3构成0,4、6构成24,其余全部和0乘即
可。详见代码:
// file name: codeforces468A.cpp // // author: kereo // // create time: 2014年09月21日 星期日 21时42分42秒 // //***********************************// #include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<set> #include<map> #include<vector> #include<stack> #include<cmath> #include<string> #include<algorithm> using namespace std; typedef long long ll; const int MAXN=100000+100; const int inf=0x3fffffff; const int mod=1000000000+7; #define L(x) (x<<1) #define R(x) (x<<1|1) int n; int main() { while(~scanf("%d",&n)){ if(n<=3){ printf("NO\n"); continue; } printf("YES\n"); if(n == 4){ printf("1 * 2 = 2\n"); printf("3 * 4 = 12\n"); printf("2 * 12 = 24\n"); continue; } if(n == 5){ printf("5 - 3 = 2\n"); printf("2 + 1 = 3\n"); printf("2 * 4 = 8\n"); printf("3 * 8 = 24\n"); continue; } printf("3 - 2 = 1\n"); printf("1 - 1 = 0\n"); printf("0 * 5 =0\n"); for(int i=7;i<=n;i++) printf("0 * %d = 0\n",i); printf("4 * 6 = 24\n"); printf("0 + 24 = 24\n"); } return 0; }
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