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| 1 | +// 多多路上从左到右有N棵树(编号1~N),其中第i个颗树有和谐值Ai。 |
| 2 | +// 多多鸡认为,如果一段连续的树,它们的和谐值之和可以被M整除,那么这个区间整体看起来就是和谐的。 |
| 3 | +// 现在多多鸡想请你帮忙计算一下,满足和谐条件的区间的数量。 |
| 4 | + |
| 5 | +// 首先注意 要求取整,那么能进行 %取余 的地方就要取余,防止精度丢失 |
| 6 | +// 例如在 readline 拿到数组之后就 map 进行 % : const nums = (await readline()).split(' ').map(item => { return item % M }) |
| 7 | + |
| 8 | +// 想要取整,本质就是区间%0,那么就是两头的前缀和在%M后相等 => 相减后为0 |
| 9 | +const rl = require("readline").createInterface({ input: process.stdin }); |
| 10 | +var iter = rl[Symbol.asyncIterator](); |
| 11 | +const readline = async () => (await iter.next()).value; |
| 12 | + |
| 13 | +void async function () { |
| 14 | + // Write your code here |
| 15 | + const [n, M] = (await readline()).split(' ') |
| 16 | + const nums = (await readline()).split(' ').map(item => { return item % M }) |
| 17 | + |
| 18 | + let cur = 0 |
| 19 | + // const pres = [0] |
| 20 | + const pres = new Map() |
| 21 | + pres.set(0, 1) // 空前缀 |
| 22 | + let ans = 0 |
| 23 | + if (!nums.length) { |
| 24 | + console.log(0) |
| 25 | + return |
| 26 | + } |
| 27 | + for (const num of nums) { |
| 28 | + cur = (cur + num) % M |
| 29 | + const pre = pres.get(cur) || 0 |
| 30 | + ans += pre |
| 31 | + pres.set(cur, pre + 1) |
| 32 | + } |
| 33 | + console.log(ans) |
| 34 | +}() |
| 35 | + |
| 36 | + |
| 37 | +// 6/10 在遍历过程中进行统计 |
| 38 | +const rl = require("readline").createInterface({ input: process.stdin }); |
| 39 | +var iter = rl[Symbol.asyncIterator](); |
| 40 | +const readline = async () => (await iter.next()).value; |
| 41 | + |
| 42 | +void async function () { |
| 43 | + // Write your code here |
| 44 | + const [n, M] = (await readline()).split(' ') |
| 45 | + const nums = (await readline()).split(' ').map(item => { return item % M }) |
| 46 | + // .map取余 |
| 47 | + |
| 48 | + let cur = 0 |
| 49 | + const pres = [0] |
| 50 | + let ans = 0 |
| 51 | + if (!nums.length) { |
| 52 | + console.log(0) |
| 53 | + return |
| 54 | + } |
| 55 | + for (const num of nums) { |
| 56 | + cur = (cur + num) % M |
| 57 | + for (const pre of pres) { |
| 58 | + if ((cur - pre) % M == 0) |
| 59 | + ans += 1 |
| 60 | + } |
| 61 | + pres.push(cur) |
| 62 | + } |
| 63 | + console.log(ans) |
| 64 | +}() |
| 65 | + |
| 66 | + |
| 67 | +// 1/10 没有在遍历过程中就进行统计 |
| 68 | +const rl = require("readline").createInterface({ input: process.stdin }); |
| 69 | +var iter = rl[Symbol.asyncIterator](); |
| 70 | +const readline = async () => (await iter.next()).value; |
| 71 | + |
| 72 | +void async function () { |
| 73 | + // Write your code here |
| 74 | + const [n, M] = (await readline()).split(' ') |
| 75 | + const nums = (await readline()).split(' ').map(item => { return item % M }) |
| 76 | + // console.log(nums) |
| 77 | + if (!nums.length) { |
| 78 | + console.log(0) |
| 79 | + return |
| 80 | + } |
| 81 | + const pre = [0] |
| 82 | + for (let i = 0; i < n; i++) { |
| 83 | + pre.push((pre[pre.length - 1] + nums[i]) % M) |
| 84 | + // 要求整除,那么处理的时候就可以取余 |
| 85 | + } |
| 86 | + let ans = 0 |
| 87 | + for (let i = 0; i < n; i++) { |
| 88 | + for (let j = i + 1; j < n + 1; j++) { |
| 89 | + if ((pre[j] - pre[i]) % M == 0) |
| 90 | + ans += 1 |
| 91 | + } |
| 92 | + } |
| 93 | + console.log(ans) |
| 94 | +}() |
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