PAT (Basic Level) Practice （中文） 1010 一元多项式求导
1 #include<stdio.h> 2 #define MAXN 100000 3 struct poly{ 4 int exp,var; 5 }; 6 struct poly a[MAXN]; 7 void read(); 8 void deal(); 9 void print(); 10 int main(){ 11 read(); 12 deal(); 13 print(); 14 return 0; 15 } 16 void read(){ 17 int e,v,count=0; 18 19 while(scanf("%d %d",&v,&e)!=EOF){ 20 a[++count].exp = e; 21 a[count].var = v; 22 } 23 a[0].exp = count; 24 a[0].var = count; 25 return; 26 27 } 28 void deal(){ 29 int e,v,ne,nv; 30 for(int i=1;i<=a[0].exp;i++){ 31 e = a[i].exp; 32 v = a[i].var; 33 nv = e * v; 34 ne = e-1; 35 a[i].exp = ne; 36 a[i].var = nv; 37 } 38 39 //合并同类 40 for(int i=1;i<=a[0].exp;i++){ 41 for(int j=i+1;j<=a[0].exp;j++){ 42 if(a[i].exp==a[j].exp){ 43 a[i].var += a[j].var; 44 a[j].var = 0; 45 a[j].exp = 0; 46 } 47 else if (a[i].exp>a[j].exp) break; 48 49 } 50 } 51 } 52 void print(){ 53 int count=0; 54 for(int i=1;i<=a[0].exp;i++) 55 if(a[i].var!=0) count++; 56 if(count==0) {printf("0 0\n");return;} 57 for(int i=1;i<=a[0].exp;i++){ 58 if(a[i].var!=0){ 59 printf("%d %d",a[i].var,a[i].exp); 60 count--; 61 if(count!=0) printf(" "); 62 } 63 } 64 65 66 }
windows vs 模拟EOF ctrl + z
