杭电acm部分题目及答案答案

自己刷的题这是我在杭电做题的记录，希望我的分享对你有帮助！！！1001 Sum Problem***********************************************************1 1089 A+B for Input-Output Practice (I)********************************21090 A+B for Input-Output Practice (II)********************************51091A+B for Input-Output Practice (III)****************************************7 1092A+B for Input-Output Practice (IV)********************************81093 A+B for Input-Output Practice (V)********************************101094 A+B for Input-Output Practice (VI)***************************************12 1095A+B for Input-Output Practice (VII)*******************************131096 A+B for Input-Output Practice (VIII)******************************15How to Type***************************************************************161001 Sum ProblemProblem DescriptionHey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.InputThe input will consist of a series of integers n, one integer per line.OutputFor each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.Sample Input1100Sample Output15050AuthorDOOM III解答：#include<stdio.h>main(){int n,i,sum;sum=0;while((scanf("%d",&n)!=-1)){sum=0;for(i=0;i<=n;i++)sum+=i;printf("%d\n\n",sum);}}1089 A+B for Input-Output Practice(I)Problem DescriptionYour task is to Calculate a + b.Too easy?! Of course! I specially designed the problem for acm beginners.You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim.InputThe input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input1 510 20Sample Output630AuthorlcyRecommendJGShining解答：#include<stdio.h>main(){int a,b;while(scanf("%d%d",&a,&b)!=EOF)printf("%d\n",a+b);}1090 A+B for Input-Output Practice(II)Problem DescriptionYour task is to Calculate a + b.InputInput contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input21 510 20Sample Output630AuthorlcyRecommendJGShining解答：#include<stdio.h>#define M 1000void main(){int a ,b,n,j[M],i;//printf("please input n:\n"); scanf("%d",&n);for(i=0;i<n;i++){scanf("%d%d",&a,&b);//printf("%d %d",a,b);j[i]=a+b;}i=0;while(i<n){printf("%d",j[i]);i++;printf("\n");}}1091A+B for Input-Output Practice(III)Problem DescriptionYour task is to Calculate a + b.InputInput contains multiple test cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input1 510 200 0Sample Output630AuthorlcyRecommendJGShining解答：#include<stdio.h>main(){int a,b;scanf("%d %d",&a,&b);while(!(a==0&&b==0)){printf("%d\n",a+b);scanf("%d %d",&a,&b);}}1092A+B for Input-Output Practice(IV)Problem DescriptionYour task is to Calculate the sum of some integers.InputInput contains multiple test cases. Each test case contains a integer N, and then N integers followin the same line. A test case starting with 0 terminates the input and this test case is not to be processed.OutputFor each group of input integers you should output their sum in one line, and with one line of output for each line in input.Sample Input4 1 2 3 45 1 2 3 4 5Sample Output1015AuthorlcyRecommendJGShining解答：#include <stdio.h>int main(){int n,sum,i,t;while(scanf("%d",&n)!=EOF&&n!=0) {sum=0;for(i=0;i<n;i++){scanf("%d",&t);sum=sum+t;}printf("%d\n",sum);}}1093 A+B for Input-Output Practice(V)Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.OutputFor each group of input integers you should output their sum in one line, and with one line of output for each line in input.Sample Input24 1 2 3 45 1 2 3 4 5Sample Output1015Authorlcy解答：#include<stdio.h>main(){int n,a,b,i,j,sum;sum=0;while(scanf("%d\n",&n)!=-1) {for(i=0;i<n;i++){scanf("%d",&b);for(j=0;j<b;j++){scanf("%d",&a);sum+=a;}printf("%d\n",sum); sum=0;}}}1094 A+B for Input-Output Practice (VI) Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains multiple test cases, and one case one line. Each case starts with an integer N, and then N integers follow in the same line.OutputFor each test case you should output the sum of N integers in one line, and with one line of output for each line in input.Sample Input4 1 2 3 45 1 2 3 4 5Sample Output1015AuthorlcyRecommendJGShining解答：#include<stdio.h>main(){int n,a,b,i,j,sum;sum=0;while(scanf("%d\n",&n)!=-1){for(j=0;j<n;j++){scanf("%d",&a);sum+=a;}printf("%d\n",sum);sum=0;}}[ Copy to Clipboard ][ Save to File]1095A+B for Input-Output Practice (VII) Problem DescriptionYour task is to Calculate a + b.InputThe input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line. OutputFor each pair of input integers a and b you should output the sum of a and b, and followed by a blank line.Sample Input1 510 20Sample Output630AuthorlcyRecommendJGShining解答：#include<stdio.h>main(){int a,b;while(scanf("%d%d",&a,&b)!=EOF)printf("%d\n\n",a+b);}1096 A+B for Input-Output Practice (VIII) Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.OutputFor each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.Sample Input34 1 2 3 45 1 2 3 4 53 1 2 3Sample Output10156AuthorlcyRecommendJGShining解答：int main(){int a,b,i,j,l[1000],k;scanf("%d",&i);getchar();for(j=1;j<=i;j++)l[j]=0;for(j=1;j<=i;j++){scanf("%d",&a);getchar();for(k=1;k<=a;k++){scanf("%d",&b);getchar();l[j]+=b;}}for(j=1;j<=i-1;j++)printf("%d\n\n",l[j]);printf("%d\n",l[i]);}How to TypeProblem DescriptionPirates have finished developing the typing software. He called Cathy to test his typing software. She is good at thinking. After testing for several days, she finds that if she types a string by some ways, she will type the key at least. But she has a bad habit that if the caps lock is on, she must turn off it, after she finishes typing. Now she wants to know the smallest times of typing the key to finish typing a string.InputThe first line is an integer t (t<=100), which is the number of test case in the input file. For each test case, there is only one string which consists of lowercase letter and upper case letter. The length of the string is at most 100.OutputFor each test case, you must output the smallest times of typing the key to finish typing this string.Sample Input 3Pirates HDUacm HDUACM Sample Output 888#include <stdio.h>#include <string.h>#define MAX 200int arr[MAX][4];char str[MAX];int letter(char ch){if(ch>='A'&&ch<='Z') return 1;return 0;}void proc(){int i;int tmp,min;int len=strlen(str);for(i=0;i<len;i++){if(i==0){if(letter(str[i])) { arr[i][1]=2; arr[i][2]=2; }else { arr[i][0]=1; arr[i][3]=3; }}else{if(letter(str[i])==letter(str[i-1])){if(arr[i-1][0]){ arr[i][0]=arr[i-1][0]+1;arr[i][3]=arr[i-1][0]+3;}if(arr[i-1][1]) { arr[i][1]=arr[i-1][1]+2; arr[i][2]=arr[i-1][1]+2;}if(arr[i-1][2]){if(arr[i][0]>arr[i-1][2]+1||!arr[i][0]) arr[i][0]=arr[i-1][2]+1;if(arr[i][3]>arr[i-1][2]+3||!arr[i][3]) arr[i][3]=arr[i-1][2]+3;}if(arr[i-1][3]){if(arr[i][1]>arr[i-1][3]+2||!arr[i][1]) arr[i][1]=arr[i-1][3]+2;if(arr[i][2]>arr[i-1][3]+2||!arr[i][2]) arr[i][2]=arr[i-1][3]+2;}}else{if(arr[i-1][0]){ arr[i][1]=arr[i-1][0]+2; arr[i][2]=arr[i-1][0]+2;}if(arr[i-1][1]){ arr[i][0]=arr[i-1][1]+1; arr[i][3]=arr[i-1][1]+3;}if(arr[i-1][2]){if(arr[i][1]>arr[i-1][2]+2||!arr[i][1]) arr[i][1]=arr[i-1][2]+2;if(arr[i][2]>arr[i-1][2]+2||!arr[i][2]) arr[i][2]=arr[i-1][2]+2;}if(arr[i-1][3]){if(arr[i][0]>arr[i-1][3]+1||!arr[i][0]) arr[i][0]=arr[i-1][3]+1;if(arr[i][3]>arr[i-1][3]+3||!arr[i][3]) arr[i][3]=arr[i-1][3]+3;}}}}min=3*MAX;if(letter(str[len-1])){if(arr[len-1][0]){ tmp=arr[len-1][0]+1; if(tmp<min) min=tmp;}if(arr[len-1][1]){ tmp=arr[len-1][1]; if(tmp<min) min=tmp; }if(arr[len-1][2]){ tmp=arr[len-1][2]+1; if(tmp<min) min=tmp;}if(arr[len-1][3]){ tmp=arr[len-1][3]; if(tmp<min) min=tmp; } }else{if(arr[len-1][0]) { tmp=arr[len-1][0]; if(tmp<min) min=tmp; }if(arr[len-1][1]) { tmp=arr[len-1][1]+1; if(tmp<min) min=tmp;}if(arr[len-1][2]) { tmp=arr[len-1][2]; if(tmp<min) min=tmp; }if(arr[len-1][3]) { tmp=arr[len-1][3]+1; if(tmp<min) min=tmp;} }printf("%d\n",min);}//Caps Shift:0-00;1-01;2-10;3-11int main(){int num;scanf("%d",&num);while(num--){scanf("%s",str);memset(arr,0,strlen(str)*4*sizeof(int));proc();}return 0;}。