Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22 Accepted Submission(s): 12Problem DescriptionLittle Ruins is a studious boy, recently he learned the four operations!
Now he want to use four operations to generate a number, he takes a string which only contains digits '1' - '9', and split it into 5 intervals and add the four operations'+', '-', '*' and '/' in order, then calculate the result(/ used as integer division).
Now please help him to get the largest result.
InputFirst line contains an integer 20
OutputFor every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the result.
Sample Input1 12345
Sample OutputCase #1: 1
Source
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题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5938
题目大意：
给一串长度为5~20的数字串（只含1~9），要求按顺序将 + - * / 各1个分别插入数字之间，使得最终运算结果最大。
题目思路：
【贪心】
分析*，因为前面是-号所以希望乘的数小，所以乘和- /之间只隔1位。
分析/，除只可能除1位数或2位数（因为乘只乘1位，3位不优，2位可能是111991，原先我以为只要/一位）。
分析+，可以知道位数越大加的结果越大，所以+只能是1位数与多位数相加。
所以综上可以分类讨论，/1位或2位，乘紧挨/前，-紧挨*，之前的数字要么在第一个断开，要么最后一个。
答案取最大即可。
1 // 2 //by coolxxx 3 //#include<bits/stdc++.h> 4 #include<iostream> 5 #include<algorithm> 6 #include<string> 7 #include<iomanip> 8 #include<map> 9 #include<stack> 10 #include<queue> 11 #include<set> 12 #include<bitset> 13 #include<memory.h> 14 #include<time.h> 15 #include<stdio.h> 16 #include<stdlib.h> 17 #include<string.h> 18 //#include<stdbool.h> 19 #include<math.h> 20 #pragma comment(linker,"/STACK:1024000000,1024000000") 21 #define min(a,b) ((a)<(b)?(a):(b)) 22 #define max(a,b) ((a)>(b)?(a):(b)) 23 #define abs(a) ((a)>0?(a):(-(a))) 24 #define lowbit(a) (a&(-a)) 25 #define sqr(a) ((a)*(a)) 26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b)) 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define eps (1e-8) 29 #define J 10000 30 #define mod 1000000007 31 #define MAX 0x7f7f7f7f 32 #define PI 3.14159265358979323 33 #define N 24 34 using namespace std; 35 typedef long long LL; 36 typedef unsigned long long ULL; 37 double anss; 38 LL aans; 39 int cas,cass; 40 int n,m,lll,ans; 41 LL sum,tot,s1,s2; 42 char s[N]; 43 int main() 44 { 45 #ifndef ONLINE_JUDGEW 46 // freopen("1.txt","r",stdin); 47 // freopen("2.txt","w",stdout); 48 #endif 49 int i,j,k; 50 int x,y,z; 51 // init(); 52 // for(scanf("%d",&cass);cass;cass--) 53 for(scanf("%d",&cas),cass=1;cass<=cas;cass++) 54 // while(~scanf("%s",s)) 55 // while(~scanf("%d%d",&n,&m)) 56 { 57 tot=0;sum=0;s1=0;s2=0; 58 printf("Case #%d: ",cass); 59 scanf("%s",s); 60 n=strlen(s); 61 tot=-(s[n-3]-'0')*(s[n-2]-'0')/(s[n-1]-'0'); 62 s1=s[0]-'0';s2=0; 63 for(i=1;i<n-3;i++) 64 s2=s2*10+s[i]-'0'; 65 sum=tot+s1+s2; 66 s1=s[n-4]-'0';s2=0; 67 for(i=0;i<n-4;i++) 68 s2=s2*10+s[i]-'0'; 69 tot+=s1+s2; 70 sum=max(sum,tot); 71 if(n>5) 72 { 73 tot=-(s[n-4]-'0')*(s[n-3]-'0')/((s[n-2]-'0')*10+s[n-1]-'0'); 74 s1=s[0]-'0';s2=0; 75 for(i=1;i<n-4;i++) 76 s2=s2*10+s[i]-'0'; 77 tot+=s1+s2; 78 sum=max(sum,tot); 79 tot=-(s[n-4]-'0')*(s[n-3]-'0')/((s[n-2]-'0')*10+s[n-1]-'0'); 80 s1=s[n-5]-'0';s2=0; 81 for(i=0;i<n-5;i++) 82 s2=s2*10+s[i]-'0'; 83 tot+=s1+s2; 84 sum=max(sum,tot); 85 } 86 printf("%lld\n",sum); 87 } 88 return 0; 89 } 90 /* 91 // 92 93 // 94 */