作业七-ID-1052-Problem D: Sequence Problem (III) : Array Practice
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Problem D: Sequence Problem (III) : Array Practice
题目描述:
解题思路:
依然要注意奇偶情况以及最后一次的情况。
与前两题不同,这一个题是每一次输出的数组规则是:
对输入的整数序列两两相加:第1行和第2行相加、第2行和第3行相加……按顺序输出结果:每行输出一个整数序列,每两个整数之间用一个空格分隔。若最后序列不足两个,则视作补一个长度为0的整数序列相加。
值得注意的是一个长度为0的整数序列也应该有输出,即使没有整数输出,也应该占有一行,因为“每行输出一个整数序列”。
给出代码:
#include <stdio.h>#include <stdlib.h>#include <string.h>int main(){ int a[2100],b[2100]; int m,n,i,j,k,x1=0,x2=0,h; scanf("%d",&k); int count1=1; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); while(count1<=k){ if(count1==1&&k==1) { scanf("%d",&j); if(j==0) { printf("\n"); } else { for(i=1;i<=j;i++) { scanf("%d",&a[i]); } x1=i-1; for(i=1;i<=x1;i++) { if(i==1) printf("%d",a[i]+b[i]); else printf(" %d",a[i]+b[i]); } printf("\n"); } break; } if(count1==1) { scanf("%d",&h); if(h==0) { x1=0; } else { for(i=1;i<=h;i++) { scanf("%d",&a[i]); } x1=i-1; } count1++; } if(count1!=1&&count1%2!=0&&count1<=k) { scanf("%d",&h); if(h==0) { x1=0; } else { for(i=1;i<=h;i++) { scanf("%d",&a[i]); } x1=i-1; } if(x1==0&&x2==0) printf("\n"); else { if(x1>x2) { for(i=1;i<=x1;i++) { if(i==1) printf("%d",a[i]+b[i]); else printf(" %d",a[i]+b[i]); } } else { for(i=1;i<=x2;i++) { if(i==1) printf("%d",a[i]+b[i]); else printf(" %d",a[i]+b[i]); } } printf("\n"); } x2=0; memset(b,0,sizeof(b)); if(count1==k) { if(x1==0) printf("\n"); else { for(i=1;i<=x1;i++) { if(i==1) printf("%d",a[i]+b[i]); else printf(" %d",a[i]+b[i]); } printf("\n"); } } count1++; } if(count1%2==0&&count1<=k) { scanf("%d",&h); if(h==0) x2=0; else { for(i=1;i<=h;i++) { scanf("%d",&b[i]); } x2=i-1; } if(x1==0&&x2==0) printf("\n"); else { if(x1>x2) { for(i=1;i<=x1;i++) { if(i==1) printf("%d",a[i]+b[i]); else printf(" %d",a[i]+b[i]); } } else { for(i=1;i<=x2;i++) { if(i==1) printf("%d",a[i]+b[i]); else printf(" %d",a[i]+b[i]); } } printf("\n"); } x1=0; memset(a,0,sizeof(a)); if(count1==k) { if(x2==0) printf("\n"); else { for(i=1;i<=x2;i++) { if(i==1) printf("%d",a[i]+b[i]); else printf(" %d",a[i]+b[i]); } printf("\n"); } } count1++; } } return 0;}
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