2143번: 두 배열의 합 
문제 링크: 2143번: 두 배열의 합
전체 코드
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#include <stdio.h>
#include <stdlib.h>
#pragma warning(disable:4996)
#pragma warning(disable:6031)
#define MAX_LEN 1000
int cmp(const void *a, const void *b);
void inputArr(int *arr, int len);
void makePrtlSum(int *orgnArr, int *newArr, int len, int *pIdx);
int main() {
int s, len1, len2, arr1[MAX_LEN + 1] = { 0 }, arr2[MAX_LEN + 1] = { 0 };
int prtlSum1[MAX_LEN * MAX_LEN], prtlSum2[MAX_LEN * MAX_LEN], sumIdx1 = 0, sumIdx2 = 0;
long long cnt = 0;
scanf("%d %d", &s, &len1);
inputArr(arr1, len1);
scanf("%d", &len2);
inputArr(arr2, len2);
makePrtlSum(arr1, prtlSum1, len1, &sumIdx1);
makePrtlSum(arr2, prtlSum2, len2, &sumIdx2);
qsort(prtlSum1, sumIdx1, sizeof(int), cmp);
qsort(prtlSum2, sumIdx2, sizeof(int), cmp);
int i0 = 0, j0 = sumIdx2 - 1;
while (i0 < sumIdx1 && j0 >= 0) {
if (prtlSum1[i0] + prtlSum2[j0] == s) {
long long currCnt = 0;
int i, j;
for (i = i0; prtlSum1[i0] == prtlSum1[i] && i < sumIdx1; i++);
for (j = j0; prtlSum2[j0] == prtlSum2[j] && j >= 0; j--);
cnt += (long long)(i - i0) * (j0 - j);
i0 = i, j0 = j;
}
else if (prtlSum1[i0] + prtlSum2[j0] < s) i0++;
else j0--;
}
printf("%lld", cnt);
return 0;
}
void inputArr(int *arr, int len) {
for (int i = 1; i <= len; i++) {
scanf("%d", &arr[i]);
arr[i] += arr[i - 1];
}
return;
}
void makePrtlSum(int *orgnArr, int *newArr, int len, int *pIdx) {
for (int i = 1; i <= len; i++) {
for (int j = 0; j < i; j++) {
newArr[(*pIdx)++] = orgnArr[i] - orgnArr[j];
}
}
return;
}
int cmp(const void *a, const void *b) {
if (*(int *)a > *(int *)b) return 1;
if (*(int *)a < *(int *)b) return -1;
return 0;
}