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1002. A+B for Polynomials (25)
时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
This time, you are supposed to find A+B where A and B are two polynomials.
Input
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < ... < N2 < N1 <=1000.
Output
For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.
1 #include <iostream> 2 3 #include <iomanip> 4 5 using namespace std; 6 7 8 9 struct fun 10 11 { 12 13 bool ifdo; 14 15 double sum; 16 17 }; 18 19 20 21 int main() 22 23 { 24 25 26 27 int n1; 28 29 while(cin>>n1) 30 31 { 32 33 fun F[1001]; 34 35 int i,num,n2; 36 37 double val; 38 39 for(i=0;i<=1000;i++) 40 41 { 42 43 F[i].ifdo=false; 44 45 F[i].sum=0.0; 46 47 } 48 49 50 51 for(i=0;i<n1;i++) 52 53 { 54 55 cin>>num>>val; 56 57 F[num].ifdo=true; 58 59 F[num].sum+=val; 60 61 } 62 63 64 65 cin>>n2; 66 67 68 69 for(i=0;i<n2;i++) 70 71 { 72 73 cin>>num>>val; 74 75 F[num].ifdo=true; 76 77 F[num].sum+=val; 78 79 } 80 81 82 83 num=0; 84 85 for(i=0;i<=1000;i++) 86 87 { 88 89 if(F[i].ifdo&&F[i].sum!=0) num++; 90 91 } 92 93 94 95 cout<<num; 96 97 98 99 for(i=1000;i>=0;i--)100 101 {102 103 if(F[i].ifdo&&F[i].sum!=0) 104 105 {106 107 108 109 cout<<" "<<i<<" "<<fixed<<setprecision(1)<<F[i].sum;110 111 }112 113 }114 115 116 117 cout<<endl;118 119 }120 121 return 0;122 123 124 125 }
1002. A+B for Polynomials (25)
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