课程资源 杭电OJ1000-1099答案

/* 1002 A + B Problem II Time Limit : 1000 ms Memory Limit : 32768 K Output Limit : 1024 K VC */ //难度在于解决大数运算的方法和实际操作中 #include <stdio.h> #include <stdlib.h> #define DIV 4 #define NUMLONG 1000 #define TMAX 20 short deadd[TMAX]={0}; short flag=0; int ChangetoNum(char * pstr,short * pnum,short now) { short ca=0,cb=3; char divstr[DIV+1]; divstr[DIV]='\0'; if(*pstr=='-') { deadd[now]++; pstr++; } while(pstr[ca]!='\0') ca++; pnum[0]=ca-ca%DIV; if(pnum[0]==0)//判断是否不足一个保存单元 pnum[0]=1; else if(pnum[0]==ca)//判断是否没有零碎的保存单元 pnum[0]=(ca-ca%DIV)/DIV; else pnum[0]=(ca-ca%DIV)/DIV+1; //pnum[0]保存数字的保存长度 pnum++;; while((ca--)>=0) { if(cb<0) { *pnum=(short)atoi(divstr);//分段字符串转换为数字 pnum++; cb=3; divstr[cb--]=pstr[ca]; } else if(ca<0 && cb<3)//无法合成一整段的字符串转换 { while(cb>=0) divstr[cb--]='0'; *pnum=(short)atoi(divstr); } else divstr[cb--]=pstr[ca]; } return 0; } int Calculate(short *pnum1,short *pnum2,short *pans) { short ca,n,temp; if(pnum1[0]<pnum2[0]) n=pnum2[0]; else n=pnum1[0]; for(ca=1;ca<=n;ca++) { temp=pnum1[ca]+pnum2[ca]+pans[ca]; if(temp>=10000)//结果数字有进位 { pans[ca]=temp-10000; pans[ca+1]=1; pans[0]=ca+1; } else pans[ca]=temp; } pans[0]=pans[0]<=n?n:pans[0]; return 0; } int Calculate2(short *pnum1,short *pnum2,short *pans)//pnum1为大数 { short ca,n,temp; n=pnum1[0]; pans[0]=n; for(ca=1;ca<=n;ca++) { temp=pnum1[ca]-pnum2[ca]-pans[ca]; if(temp==0 && ca==n) pans[0]=n-1; if(temp<0)//结果数字要退位 { if(pnum1[ca+1]>0) { pans[ca]=temp+10000; pans[ca+1]=1; } } else pans[ca]=temp; } return 0; } int PrintAns(char *pleft,char *pright,short *pans,short ca) { short cb,temp; printf("Case %d:\n",ca); printf("%s + %s = ",pleft,pright); if(flag==1) //首单元输出特殊化 printf("%d",-pans[pans[0]]); else printf("%d",pans[pans[0]]); for(ca=pans[0]-1;ca>=1;ca--) { if(pans[ca]>=1000)//4位全满 printf("%d",pans[ca]); else { //if(ca<pans[0])//判断是否为第一个单元 { temp=pans[ca]; for(cb=0;temp>0;cb++)//输出数字中间的0 temp/=10; for(;cb<4;cb++) printf("0"); } if(pans[ca]!=0) printf("%d",pans[ca]);//当此单元为0时，0已经输出完毕，就不用再输出了 } } printf("\n"); return 0; } short WhoBig(short *pnum1,short *pnum2) { short ca; if(pnum1[0]>pnum2[0]) return 1; else if(pnum1[0]<pnum2[0]) return -1; for(ca=pnum1[0];ca>=1;ca--) { if(pnum1[ca]==pnum2[ca]) continue; if(pnum1[ca]>pnum2[ca]) return 1; else return -1; } return 0; } int main() { short n,ca,big; char strleft[TMAX][NUMLONG+1],strright[TMAX][NUMLONG+1]; short numleft[TMAX][NUMLONG/DIV]={0},numright[TMAX][NUMLONG/DIV]={0}; short ans[TMAX][NUMLONG/DIV]={0}; scanf("%d",&n); ca=0; while(ca<n && scanf("%s %s",strleft[ca],strright[ca])==2) { ChangetoNum(strleft[ca],numleft[ca],ca);//正正为3，正负为4，负正为1，负负为2 if(deadd[ca]==0) deadd[ca]=3; ChangetoNum(strright[ca],numright[ca],ca); ca++; }//while输入数据 for(ca=0;ca<n;ca++) { switch(deadd[ca]) { case 3: Calculate(numleft[ca],numright[ca],ans[ca]); flag=0; break; case 2: Calculate(numleft[ca],numright[ca],ans[ca]); flag=1; break; case 4: case 1: big=WhoBig(numleft[ca],numright[ca]); if(big==0) { printf("Case %d:\n",ca+1); printf("%s + %s = 0\n",strleft[ca],strright[ca]); continue; } else if(big==1)//左边的数大于右边的数 { Calculate2(numleft[ca],numright[ca],ans[ca]); if(deadd[ca]==4)//正负 flag=0; else if(deadd[ca]==1)//负正 flag=1; } else if(big==-1)//右边的数大于左边的数 { Calculate2(numright[ca],numleft[ca],ans[ca]); if(deadd[ca]==4)//正负 flag=1; else if(deadd[ca]==1)//负正 flag=0; } break; } PrintAns(strleft[ca],strright[ca],ans[ca],ca+1); if(ca!=n-1) printf("\n"); }//for处理数据 return 0; }