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/**/#include <stdio.h> #include <stdlib.h> #include <string.h> #include <ctype.h> /*int cchkdig(char *r) { int i=0; while(r[i]!='\0') { if(isdigit(r[i++])==0) return (0); } return (1); } */void del_0(char *r) //去掉整数串表示前面多余的零，最后结果为空串时置为"0" { unsigned int lr; int i=0, j; lr=strlen(r); while(r[i]=='0') ++i; if(i>0) { for(j=0; j<lr-i; ++j) r[j]=r[j+i]; for(j=lr-i; j<lr; ++j) { r[j]='\0'; } } if(r[0]=='\0') { r[0]='0'; } } int scmp(char *r, char *u) //比较长度{ unsigned int lr, lu; del_0(r); del_0(u); lr=strlen(r); lu=strlen(u); if(lr>lu) { return 1; } else if (lr<lu) { return -1; } return (strcmp(r, u)); }char *sub(char *r, char *u) //两个串表示数的减法 { unsigned int i,lr, lu, lp,c=0; char h,hc; char *p; if(scmp(r, u)<0) //r比u大 return NULL; lr=strlen(r); lu=strlen(u); p=(char *)malloc((unsigned int)(lr+1)*sizeof(char)); for(i=0; i<lu; ++i) { h=r[lr-i-1]-u[lu-i-1]-c; if(h<0) { c=1; h=h+10; } else c=0; p[i]=h+'0'; hc=h+'0'; } for (i=lu; i<lr; ++i) { h=r[lr-i-1]-'0'-c; if(h<0) { c=1; h=h+10; } else c=0; p[i]='0'+h; hc='0'+h; } p[i]='\0'; lp=i-1; while(p[lp]=='0'&&lp!=0) { p[lp]='\0'; lp--; } for(i=0; i<(lp+1)/2; ++i) { hc=p[i]; p[i]=p[lp-i]; p[lp-i]=hc; } return (p); }//end sub() char *add(char *r, char *u) //两个串表示数的加法 { unsigned int lr, lu, lp; int i, h, c=0; char hc, *p; lr=strlen(r); lu=strlen(u); if(lu>lr) { p=r; r=u; u=p; h=lr; lr=lu; lu=h; } p=(char *)malloc((unsigned int)(lr+2)*sizeof(char)); for(i=0; i<lu; ++i) { h=r[lr-i-1]-'0'+u[lu-i-1]-'0'+c; if(h>9) { c=1; h=h-10; } else c=0; p[i]=h+'0'; } for(i=lu; i<lr; ++i) { h=r[lr-i-1]-'0'+c; if(h>9) { c=1; h=h-10; } else c=0; p[i]='0'+h; } if(c>0) { p[i]=c+'0'; lp=i; } else lp=i-1; for(i=lp+1; i<lr+2; ++i) p[i]='\0'; for(i=0; i<(lp+1)/2; ++i) { hc=p[i]; p[i]=p[lp-i]; p[lp-i]=hc; } return (p); }//end add() char *mul(char *r, char *u) //两个串表示数的乘法 { unsigned int lr, lu, lp; int i, j, c, h; char *p; lr=strlen(r); lu=strlen(u); p=(char *)malloc((unsigned int)(lr+lu+1)*sizeof(char)); for(i=0; i<lr+lu; ++i) p[i]='0'; p[lr+lu]='\0'; for(i=lr-1; i>=0; --i) { c=0; for(j=lu-1; j>=0; --j) { lp=i+j+1; h=(r[i]-'0')*(u[j]-'0')+p[lp]-'0'+c; c=h/10; h=h%10; p[lp]=h+'0'; } if(c>0)p[i+j+1]=c+'0'; } del_0(p); return p; }//end mul() char *div(char *u, char *v, int n) //两个串表示数的除法，结果精确到小数点后第n位 { char *p, *f, *r,*q; unsigned int i, lu, lv, lr, iw, c, h; int kh, j; lu=strlen(u); lv=strlen(v); f=(char *)malloc((unsigned int)(lu+n+3)*sizeof(char)); q=(char *)malloc(sizeof(char)); for(i=0; i<lu+n+3; ++i) f[i]='\0'; r=(char *)malloc((unsigned int)(lv+2)*sizeof(char)); for(i=0; i<lv+2; ++i) r[i]='\0'; for(iw=0; iw<lu+n+2; ++iw) { if(iw<lu) { del_0(r); lr=strlen(r); r[lr]=u[iw]; r[lr+1]='\0'; } else if(iw>lu) { del_0(r); q[0]='0'; if(scmp(r, q)==0) { break; } lr=strlen(r); r[lr]='0'; r[lr+1]='\0'; } else { f[lu]='.'; continue; } kh=0; while(scmp(r, v)>=0) { p=r; r=sub(p, v); ++kh; } f[iw]=kh+'0'; } if(iw==lu+n+2) { if(f[lu+n+1]>='5') { f[lu+n+1]='\0'; c=1; for(j=lu+n; j>=0; --j) { if(c==0) { break; } if(f[j]=='.') { continue; } h=f[j]-'0'+c; if(h>9) { h=h-10; c=1; } else c='\0'; f[j]=h+'0'; } } else f[lu+n+1]='\0'; } free(r); free(p); q=NULL; free(q); del_0(f); return(f); }//end div() /*//两个串表示数的除法，结果分别用整商与余数表示 char *sdivkr(char *u, char *v, char **rout) { char *f, *r; unsigned int i, lu, lv, lr, iw; int kh; lu=strlen(u); lv=strlen(v); f=(char *)malloc((unsigned int)(lu+1)*sizeof(char)); for(i=0; i<lu+1; ++i) f[i]='\0'; r=(char *)malloc((unsigned int)(lv+2)*sizeof(char)); for(i=0; i<lv+2; ++i) r[i]='\0'; for(iw=0; iw<lu; ++iw) { del_0(r); lr=strlen(r); r[lr]=u[iw]; r[lr+1]='\0'; kh=0; while(scmp(r, v)>=0) { r=sub(r, v); ++kh; } f[iw]=kh+'0'; } del_0(r); *rout=r; del_0(f); return(f);}//end *sdivkr() *//*调用上述函数实现两任意长正整数任意指定精度的算术计算器程序 void main(int argc, char *argv[]) { char *p, *r; int n; if(argc!=4) { if(argc!=3) printf("\n>>\"order n1 op n2\" or n ! "); exit(0); } del_0(argv[1]); if(cchkdig(argv[1])==0) { printf("Input data error, Input again!"); exit(0); } del_0(argv[3]); if(cchkdig(argv[3])==0) { printf("Input data error, Input again!"); exit(0); } if(strcmp(argv[2], "+")==0) { printf("%s", p=add(argv[1], argv[3])); free(p); } else if(strcmp(argv[2], "-")==0) { printf("%s", p=sub(argv[1], argv[3])); free(p); } else if(strcmp(argv[2], "*")==0) { printf("%s", p=mul(argv[1], argv[3])); free(p); } else if(argv[2][0]=='/' && strlen(argv[2])==1) { if(argv[3][0]=='0') { printf("error!devided by zero!!\n"); exit(0); } p=sdivkr(argv[1], argv[3], &r); printf("k=%s r=%s", p, r); free(p); free(r); } else if(argv[2][0]=='/'&&strlen(argv[2])>1) { if(argv[3][0]=='0') { printf("error!devided by zero!!\n"); exit(0); } argv[2][0]='\0'; del_0(argv[2]); if(cchkdig(argv[2])==0) { printf("Input data error, Input again!"); exit (0); } n=atoi(argv[2]); printf("%s", p=div(argv[1], argv[3], n)); free(p); } } */char a[200000];int main(){ //freopen("read.txt", "r", stdin); int n; scanf("%d", &n); while(n--) { memset(a, '\0', sizeof(a)); char b[200] = { "0"}; while(~scanf("%s", a) && a[0] != 0) { strcpy(b, add(a, b)); } printf("%s\n", b); if(n!=0) printf("\n"); } return 0;}
转载于:https://www.cnblogs.com/sober-reflection/p/1bcdc370a0830eb0727da1c9ab766cbf.html