标签:ring return 分拆 mat ret strcpy itoa class cci
1 #include<stdio.h> 2 #include<string.h> 3 4 int main(){ 5 int n,i,r,x,j,flag; 6 int s[1000]; 7 while(scanf("%d %d",&n,&r)!=EOF){ 8 flag = 0; 9 if(n<0){flag=-1;n=-n;} 10 i = 0; 11 while(n>0){ 12 x = n%r; 13 s[i] = x; 14 i++; 15 n = n/r; 16 } 17 if(flag==-1){ 18 printf("-"); 19 } 20 for(j=i-1;j>=0;j--){ 21 if(s[j]==10){printf("A");} 22 else if(s[j]==11){printf("B");} 23 else if(s[j]==12){printf("C");} 24 else if(s[j]==13){printf("D");} 25 else if(s[j]==14){printf("E");} 26 else if(s[j]==15){printf("F");} 27 else{printf("%d",s[j]);} 28 } 29 printf("\n"); 30 } 31 }
1 #include<stdio.h> 2 #include<string.h> 3 4 int main(){ 5 int n,AH,AM,AS,BH,BM,BS,SUMH,SUMM,SUMS; 6 while(scanf("%d",&n)!=EOF){ 7 while(n--){ 8 scanf("%d %d %d %d %d %d",&AH,&AM,&AS,&BH,&BM,&BS); 9 SUMS = AS + BS; 10 SUMM = AM + BM; 11 SUMH = AH + BH; 12 if(SUMS>=60){SUMM++;SUMS -= 60;} 13 if(SUMM>=60){SUMH++;SUMM -= 60;} 14 printf("%d %d %d\n",SUMH,SUMM,SUMS); 15 } 16 } 17 18 }
1 #include<stdio.h> 2 #include<string.h> 3 4 int main(){ 5 int i,j; 6 _int64 a[10000]; 7 while(scanf("%d",&i)!=EOF){ 8 if(i==-1){break;} 9 a[0]=0;a[1]=1; 10 for(j=2;j<=i;j++){ 11 a[j] = a[j-1] + a[j-2]; 12 } 13 printf("%I64d\n",a[i]); 14 } 15 16 17 }
1 #include<stdio.h> 2 #include<string.h> 3 4 int main(){ 5 int n,m,i; 6 double h[110]; 7 while(scanf("%d",&n)!=EOF){ 8 while(n--){ 9 scanf("%d",&m); 10 for(i=1;i<=m;i++){ 11 scanf("%lf",&h[i]); 12 } 13 h[0] = 0; 14 for(i=1;i<=m;i++){ 15 if(h[0]<h[i]){ 16 h[0]=h[i]; 17 } 18 } 19 printf("%.2lf\n",h[0]); 20 21 } 22 } 23 }
1 #include<stdio.h> 2 3 int main(){ 4 _int64 a,b; 5 int n; 6 while(scanf("%d",&n)!=EOF){ 7 while(n--){ 8 scanf("%I64d %I64d",&a,&b); 9 if(a%b==0){printf("YES\n");} 10 else{printf("NO\n");} 11 } 12 } 13 }
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 5 int s[1000000]; 6 int main(){ 7 int n,m,i,count; 8 char stri[10]; 9 for(i=0;i<=1000000;i++){ 10 itoa(i,stri,10); 11 if(strstr(stri,"4")==NULL && strstr(stri,"62")==NULL){s[i]=1;} 12 else{s[i]=0;} 13 } 14 while(scanf("%d %d",&n,&m)!=EOF){ 15 if(n==0&&m==0){break;} 16 count = 0; 17 18 for(i=n;i<=m;i++){count += s[i];} 19 printf("%d\n",count); 20 } 21 22 }
这里有点坑,首先先要生成100000个数据的数组,提前判断是否是不吉利数,再输入后就直接查询,不然就会Time Limit Exceeded。
1 #include<stdio.h> 2 3 4 int main(){ 5 double n,p,sum=0; 6 char s[1000]; 7 while(scanf("%s %lf %lf",&s,&n,&p)!=EOF){ 8 9 sum += n*p; 10 } 11 printf("%.1lf\n",sum); 12 }
1 #include<stdio.h> 2 3 4 int main(){ 5 int n,m,a,k,flag; 6 while(scanf("%d %d",&n,&m)!=EOF){ 7 flag=0; 8 if(n==0&&m==0){break;} 9 if(m==0){printf("Yes\n");} 10 if(m<0){k=-m;}else{k=m;} 11 for(a=1;a<=k/2;a++){ 12 if(m%a==0 && (a+m/a==n || (-a)+m/(-a)==n)){printf("Yes\n");flag=1;break;} 13 } 14 if(flag==0){printf("No\n");} 15 } 16 }
1 #include<stdio.h> 2 3 int main(){ 4 int n,a,b,sum; 5 while(scanf("%d",&n)!=EOF){ 6 while(n--){ 7 scanf("%d %d",&a,&b); 8 sum = a%100+b%100; 9 if(sum>99){sum %= 100;} 10 printf("%d\n",sum); 11 } 12 } 13 }
1 #include<stdio.h> 2 3 int hex(int n){ 4 int sum=0; 5 while(n>15){ 6 sum = sum + n%16; 7 n /= 16; 8 } 9 sum = sum + n; 10 return sum; 11 } 12 13 int dms(int n){ 14 int sum=0; 15 while(n>11){ 16 sum = sum + n%12; 17 n /= 12; 18 } 19 sum = sum + n; 20 return sum; 21 } 22 23 int main(){ 24 int n,a,b,c,d,sum,sum1,sum2; 25 while(~scanf("%d",&n)){ 26 if(n==0){break;} 27 a = n%10; 28 b = n/10%10; 29 c = n/100%10; 30 d = n/1000; 31 sum = a+b+c+d; 32 sum1 = hex(n); 33 sum2 = dms(n); 34 if(sum==sum1 && sum==sum2){printf("%d is a Sky Number.\n",n);} 35 else{printf("%d is not a Sky Number.\n",n);} 36 } 37 }
1 #include<stdio.h> 2 #include<math.h> 3 4 int ispn(int n){ 5 int i,flag=0; 6 for(i=2;i<=n/2;i++){ 7 if(n%i==0){flag=1;break;} 8 } 9 return flag; 10 } 11 12 int main(){ 13 int n,i,count; 14 while(~scanf("%d",&n)){ 15 if(n==0){break;} 16 count = 0; 17 for(i=2;i<n/2;i++){ 18 if(ispn(i)==0 && ispn(n-i)==0){count++;} 19 } 20 printf("%d\n",count); 21 } 22 }
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 5 6 int main(){ 7 int a,b,i,j,len,k; 8 char s[100][10],str[10]; 9 while(~scanf("%d %d",&a,&b)){ 10 if(a==0&&b==0){break;} 11 j=0; 12 for(i=a*100;i<=a*100+99;i++){ 13 if(i%b==0){ 14 itoa(i,str,10); 15 strcpy(s[j],str); 16 j++; 17 } 18 } 19 for(i=0;i<j;i++){ 20 k = atoi(s[i]); 21 if(i==j-1){ 22 if(k%100<10){printf("0%d\n",k%100);} 23 else{printf("%d\n",k%100);} 24 }else{ 25 if(k%100<10){printf("0%d ",k%100);} 26 else{printf("%d ",k%100);} 27 } 28 } 29 } 30 }
标签:ring return 分拆 mat ret strcpy itoa class cci
原文地址:https://www.cnblogs.com/Ragd0ll/p/10458967.html