hdu5308 I Wanna Become A 24-Point Master(构造)

来源：互联网发布：java 进程注入编辑：程序博客网时间：2024/06/06 06:50

题目：http://acm.hdu.edu.cn/showproblem.php?pid=5308

题意：给定N个N,求出将这N个N构造成24的方法。

分析：要是想到用尽量少的N构造出24点（(N+N)/N * (N+N)/N * (N+N)/N * (N*N+N)/N），然后把多的N构造成0就好办了（(N-N)*N*.....）。

代码：

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;void _15(int n){printf("1 + 2\n");//n+1printf("4 + 5\n");//n+2printf("7 + 8\n");//n+3printf("10 + 11\n");//n+4printf("%d + 12\n",n+4);//n+5printf("%d / 3\n",n+1); // n+6 2printf("%d / 6\n",n+2);//n+7   2printf("%d / 9\n",n+3);//n+8   2printf("%d / 13\n",n+5);//n+9  3printf("14 - 15\n");//n+10     0int c=n+10;for(int i=16;i<=n;i++){printf("%d * %d\n",c,i);c++;}printf("%d * %d\n",n+6,n+7); //c+1printf("%d * %d\n",n+8,c+1); //c+1printf("%d * %d\n",c+2,n+9);printf("%d + %d\n",c+3,c);}void _14(){int s1[100]={1,15,16,17,18,7,9,11,20,21,22,25,26};int s2[100]={2,3,4,5,6,8,10,12,13,14,19,23,24};char op[100]="-****+++//+--";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _13(){int s1[100]={1,4,7,10,17,14,15,16,18,19,21,23};int s2[100]={2,5,8,11,12,3,6,9,13,20,22,24};char op[100]="+++++////***";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _12(){int s1[100]={1,3,5,7,9,11,13,15,17,19,21};int s2[100]={2,4,6,8,10,12,14,16,18,20,22};char op[100]="-----++++++";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _11(){int s1[100]={1,12,13,14,15,7,17,9,18,19};int s2[100]={2,3,4,5,6,8,16,10,11,20};char op[100]="-****+++/+";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _10(){int s1[100]={1,11,4,6,8,13,14,15,17};int s2[100]={2,3,5,7,9,10,12,16,18};char op[100]="-++++//++";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _9(){int s1[100]={1,3,4,11,6,8,15,13};int s2[100]={2,10,5,12,7,14,9,16};char op[100]="++-+++/-";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _8(){int s1[100]={1,9,10,11,12,13,14};int s2[100]={2,3,4,5,6,7,8};char op[100]="-***+++";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _7(){int s1[100]={1,3,4,6,11,12};int s2[100]={2,8,5,10,7,9};char op[100]="++++/+";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _6(){int s1[100]={1,3,7,5,9};int s2[100]={2,4,8,6,10};char op[100]="+++-+";for(int i=0;op[i];i++)printf("%d %c %d\n",s1[i],op[i],s2[i]);}void _5(){printf("1 * 2\n");printf("6 * 3\n");printf("7 - 4\n");printf("8 / 5\n");}void _4(){printf("1 * 2\n");printf("5 + 3\n");printf("6 + 4\n");}int main(){int n;while(scanf("%d",&n)!=EOF){if(n>=15)_15(n);else if(n==14)_14();else if(n==13)_13();else if(n==12)_12();else if(n==11)_11();else if(n==10)_10();else if(n==9)_9();else if(n==8)_8();else if(n==7)_7();else if(n==6)_6();else if(n==5)_5();else if(n==4)_4();elseprintf("-1\n");}return 0;}

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