百练2981 http://poj.grids.cn/practice/2981/
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1 #include<stdio.h> 2 #include<string.h> 3 int num[301]; 4 int main() 5 { 6 int i,j,f = 0,f1; 7 char c1[201],c2[201]; 8 gets(c1); 9 gets(c2); 10 i = strlen(c1)-1; 11 j = strlen(c2)-1; 12 int g = 0; 13 while(i>=0&&j>=0) 14 { 15 g++; 16 if(c1[i]-'0'+c2[j]-'0'+num[g]>9) 17 { 18 num[g+1] += (c1[i]-'0'+c2[j]-'0'+num[g])/10; 19 num[g] = (c1[i]-'0'+c2[j]-'0'+num[g])%10; 20 f1 = 0; 21 } 22 else 23 num[g] += c1[i]-'0'+c2[j]-'0'; 24 i--;j--; 25 } 26 while(i>=0) 27 { 28 g++; 29 if(num[g]+c1[i]-'0'>9) 30 { 31 num[g+1] += (num[g]+c1[i]-'0')/10; 32 num[g] = (num[g]+c1[i]-'0')%10; 33 } 34 else 35 num[g] += c1[i]-'0'; 36 i--; 37 } 38 while(j>=0) 39 { 40 g++; 41 if(num[g]+c1[j]-'0'>9) 42 { 43 num[g+1] += (num[g]+c1[j]-'0')/10; 44 num[g] = (num[g]+c1[j]-'0')%10; 45 } 46 else 47 num[g] = num[g]+c1[j]-'0'; 48 j--; 49 } 50 g++; 51 while(num[g]==0&&g) 52 g--; 53 for(i = g ; i>= 1; i--) 54 { 55 f = 1; 56 printf("%d",num[i]); 57 } 58 if(!f) 59 printf("0"); 60 printf("\n"); 61 return 0; 62 }
hdu 1002 http://acm.hdu.edu.cn/showproblem.php?pid=1002
加法 先字符串化整 直接相加 再进位
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1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 int i,j,k,t,a1[1001],a2[1001],g,f; 6 char c1[1001],c2[1001]; 7 scanf("%d%*c",&t); 8 for(j = 1; j <= t ; j++) 9 { 10 scanf("%s %s",c1,c2); 11 memset(a1,0,sizeof(a1)); 12 memset(a2,0,sizeof(a2)); 13 f = 0; 14 for(i = 0 ; i < strlen(c1) ; i++) 15 a1[strlen(c1)-i-1] = c1[i]-'0'; 16 for(i = 0 ; i < strlen(c2) ; i++) 17 a2[strlen(c2)-i-1] = c2[i]-'0'; 18 if(strlen(c1)>strlen(c2)) 19 k = strlen(c1); 20 else 21 k = strlen(c2); 22 for(i = 0 ; i < k ; i++) 23 a2[i] += a1[i]; 24 for(i = 0 ; i < k ; i++) 25 if(a2[i]>9) 26 { 27 a2[i+1] += a2[i]/10; 28 a2[i] = a2[i]%10; 29 } 30 printf("Case %d:\n",j); 31 printf("%s + %s = ",c1,c2); 32 while(a2[i]==0&&i>=0) 33 i--; 34 for(g = i ; g >= 0 ; g--) 35 { 36 f = 1; 37 printf("%d",a2[g]); 38 } 39 if(!f) 40 printf("0"); 41 printf("\n"); 42 if(j!=t) 43 printf("\n"); 44 } 45 return 0; 46 }
hdu1047 http://acm.hdu.edu.cn/showproblem.php?pid=1047
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1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 int i,j,t,num[101][201],toal[201],g; 6 char c[101][101]; 7 scanf("%d%*c", &t); 8 while(t--) 9 { 10 memset(num,0,sizeof(num)); 11 memset(toal,0,sizeof(toal)); 12 i = 0; 13 int f = 0; 14 while(gets(c[i])!=NULL) 15 { 16 if(strcmp(c[i],"0")==0) 17 break; 18 i++; 19 } 20 int max = 0; 21 for(j = 0 ; j < i ; j++) 22 { 23 for(g = 0 ; g < strlen(c[j]) ; g++) 24 num[j][strlen(c[j])-g-1] = c[j][g]-'0'; 25 if(strlen(c[j])>max) 26 max = strlen(c[j]); 27 } 28 for(g = 0 ; g < max ; g++) 29 for(j = 0 ; j < i ; j++) 30 toal[g]+=num[j][g]; 31 for(g = 0 ; g < max+3 ; g++) 32 if(toal[g]>9) 33 { 34 toal[g+1] += toal[g]/10; 35 toal[g] = toal[g]%10; 36 } 37 while(toal[g]==0&&g>=0) 38 g--; 39 for(i = g ; i>=0 ; i--) 40 { 41 f = 1; 42 printf("%d",toal[i]); 43 } 44 if(!f) 45 printf("0"); 46 printf("\n"); 47 if(t!=0) 48 puts(""); 49 } 50 return 0; 51 }
hdu 1753 http://acm.hdu.edu.cn/showproblem.php?pid=1753
小数的加法 化为整数的加法 注意小数位不一样时 少的补0 前导0和小数后面多余的0要去掉
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1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 int i,j,k,t,a1[1001],a2[1001],g,f,k1,k2,x,y,w; 6 char c1[1001],c2[1001]; 7 while(scanf("%s %s",c1,c2)!=EOF) 8 { 9 memset(a1,0,sizeof(a1)); 10 memset(a2,0,sizeof(a2)); 11 k1 = strlen(c1); 12 k2 = strlen(c2); 13 w =0;f = 0;x=y=0; 14 for(i = 0 ; i < k1 ; i++) 15 if(c1[i]=='.') 16 { 17 w = 1; 18 x = k1-1-i; 19 for(j = i ; j < k1-1 ; j++) 20 c1[j] = c1[j+1]; 21 break; 22 } 23 if(w) 24 k1--; 25 w = 0; 26 for(i = 0 ; i < k2 ; i++) 27 if(c2[i]=='.') 28 { 29 w = 1; 30 y = k2-1-i; 31 for(j = i ; j < k2-1 ; j++) 32 c2[j] = c2[j+1]; 33 break; 34 } 35 if(w) 36 k2--; 37 int g1=0,g2=0; 38 if(x<y) 39 { 40 for(i = 0 ; i < y-x ; i++) 41 a1[g1++] = 0; 42 x = y; 43 } 44 else 45 for(i = 0 ; i < x-y ; i++) 46 a2[g2++] = 0; 47 for(i = k1-1 ; i >= 0 ; i--) 48 a1[g1++] = c1[i]-'0'; 49 for(i = k2-1 ; i >= 0 ; i--) 50 a2[g2++] = c2[i]-'0'; 51 if(g1>g2) 52 g2 = g1; 53 for(i = 0 ; i < g2 ; i++) 54 a2[i] += a1[i]; 55 for(i = 0 ; i < g2 ; i++) 56 if(a2[i]>9) 57 { 58 a2[i+1] += a2[i]/10; 59 a2[i] = a2[i]%10; 60 } 61 while(a2[i]==0&&i>=0) 62 i--; 63 j = 0; 64 while(a2[j]==0&&j<x) 65 j++; 66 for(g = i ; g >= j ; g--) 67 { 68 f = 1; 69 if(g+1==x) 70 { 71 if(g==i) 72 printf("0"); 73 printf("."); 74 } 75 printf("%d",a2[g]); 76 } 77 if(!f) 78 printf("0"); 79 printf("\n"); 80 } 81 return 0; 82 }
百练2736 http://poj.grids.cn/practice/2736/
减法 直接把第二个数都变成负数相加
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1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 int i,j,k,t,a1[1001],a2[1001],g,f; 6 char c1[1001],c2[1001]; 7 scanf("%d%*c",&t); 8 for(j = 1; j <= t ; j++) 9 { 10 scanf("%s %s",c1,c2); 11 memset(a1,0,sizeof(a1)); 12 memset(a2,0,sizeof(a2)); 13 f = 0; 14 for(i = 0 ; i < strlen(c1) ; i++) 15 a1[strlen(c1)-i-1] = c1[i]-'0'; 16 for(i = 0 ; i < strlen(c2) ; i++) 17 a2[strlen(c2)-i-1] = (c2[i]-'0')*-1; 18 if(strlen(c1)>strlen(c2)) 19 k = strlen(c1); 20 else 21 k = strlen(c2); 22 for(i = 0 ; i < k ; i++) 23 a1[i] += a2[i]; 24 for(i = 0 ; i < k ; i++) 25 if(a1[i]<0) 26 { 27 a1[i+1]--; 28 a1[i] = a1[i]+10; 29 } 30 while(a1[i]==0&&i>=0) 31 i--; 32 for(g = i ; g >= 0 ; g--) 33 { 34 f = 1; 35 printf("%d",a1[g]); 36 } 37 if(!f) 38 printf("0"); 39 printf("\n"); 40 } 41 return 0; 42 }
http://poj.org/problem?id=1001
化为整数相乘 再算出小数点在哪里
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1 #include<stdio.h> 2 #include<string.h> 3 int num[1001]; 4 int main() 5 { 6 int i,j,k,n,x,y,w,o,g,a; 7 char c[15],c1[1001],c2[1001]; 8 while(scanf("%s %d", c,&n)!=EOF) 9 { 10 memset(num,0,sizeof(num)); 11 memset(c2,'0',sizeof(c2)); 12 k = strlen(c); 13 x = 0; 14 int f = 0; 15 for(i = 0 ; i < k ;i++) 16 if(c[i] == '.') 17 { 18 x = k-1-i; 19 for(j = i ; j < k-1 ; j++) 20 c[j] = c[j+1]; 21 k--; 22 break; 23 } 24 c[k] = '\0'; 25 o = 0; 26 x = n*x; 27 for(i = k-1 ; i >= 0 ; i--) 28 { 29 30 if(n==1) 31 c2[o++] = c[i]; 32 else 33 c1[o++] = c[i]; 34 } 35 for(i = 1; i < n ; i++) 36 { 37 w = 0; 38 memset(c2,'0',sizeof(c2)); 39 for(j = k-1 ; j>=0 ; j--) 40 { 41 g = w; 42 for(y = 0 ; y< o ; y++) 43 { 44 if(c2[g]-'0'+(c[j]-'0')*(c1[y]-'0')>9) 45 { 46 c2[g+1]+=(c2[g]-'0'+(c[j]-'0')*(c1[y]-'0'))/10; 47 c2[g] = (c2[g]-'0'+(c[j]-'0')*(c1[y]-'0'))%10+'0'; 48 } 49 else 50 c2[g]+= (c[j]-'0')*(c1[y]-'0'); 51 g++; 52 } 53 w++; 54 } 55 if(c2[g]=='0') 56 o = g; 57 else 58 o = g+1; 59 for(a = 0 ; a < o ; a++) 60 c1[a] = c2[a]; 61 } 62 o++; 63 while(c2[o]=='0'&&o>=x) 64 o--; 65 j = 0; 66 while(c2[j]=='0'&&j<x) 67 j++; 68 for(i = o ; i >= j ; i--) 69 { 70 f = 1; 71 if(i+1==x) 72 printf("."); 73 printf("%c",c2[i]); 74 } 75 if(!f) 76 printf("0"); 77 printf("\n"); 78 } 79 return 0; 80 }